# What is the pH after mixing the following?

## a) 12.50 mL of 0.200 mol/L hydrochloric acid and 15.00 mL of 0.100 mol/L calcium hydroxide b) 3.50 g of sodium sulfate in 1.50 L of water c) 10.00 mL of 0.200 mol/L ammonia and 10.00 mL of 0.2 mol/L ammonium chloride d) 10.00 mL of 0.200 mol/L sodium hydroxide and 25.00 mL of 0.200 mol/L acetic acid e) 30 mL of 0.1 mol/L potassium hydroxide and 15 mL of 0.2 mol/L hydrocyanic acid f) 30.00 mL of 0.100 mol/L hydrochloric acid and and 25.00 mL of 0.100 mol/L ammonia

Sep 27, 2016

For part (a): $\text{pH} = 12.260$

#### Explanation:

I'm going to start this off by solving part (a).

In this case, you're mixing hydrochloric acid, $\text{HCl}$, a strong acid, and calcium hydroxide, "Ca"("OH")_2, a strong base.

You are told that all the base dissolves, which means that the solution contains twice as many moles of hydroxide anions, ${\text{OH}}^{-}$, as moles of calcium hydroxide used to make the solution.

${\text{Ca"("OH")_ (color(red)(2)(s)) -> "Ca"_ ((aq))^(2+) + color(red)(2)"OH}}_{\left(a q\right)}^{-}$

Use the molarities and volumes of the two solutions to find how many moles of each you're mixing

12.50 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.200 moles HCl"/(1color(red)(cancel(color(black)("L")))) = "0.00250 moles HCl"

15.00 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.100 moles Ca"("OH")_2)/(1color(red)(cancel(color(black)("L")))) = "0.00150 moles Ca"("OH")_2

The number of moles of hydroxide anions delivered to the solution will thus be

0.00150 color(red)(cancel(color(black)("moles Ca"("OH")_2))) * (color(red)(2)color(white)(a)"moles OH"^(-))/(1color(red)(cancel(color(black)("mole Ca"("OH")_2)))) = "0.00300 moles OH"^(-)

Now, the balanced chemical equation that describes this neutralization reaction looks like this

$2 {\text{HCl"_ ((aq)) + "Ca"("OH")_ (2(s)) -> "CaCl"_ (2(aq)) + 2"H"_ 2"O}}_{\left(l\right)}$

Because you assume that all the base is dissolved, you can write the net ionic equation like this

$2 {\text{H"_ ((aq))^(+) + 2"OH"_ ((aq))^(-) -> 2"H"_ 2"O}}_{\left(l\right)}$

which, of course, gets you

${\text{H"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "H"_ 2"O}}_{\left(l\right)}$

So, the reaction consumes equal numbers of moles of hydrogen ions, ${\text{H}}^{+}$, and hydroxide anions. Because you have fewer moles of hydrogen ions coming from the hydrochloric acid than moles of hydroxide anions coming from the calcium hydroxide, the hydrochloric acid will be a limiting reagent.

In other words, the hydrogen ions coming from the acid will be completely consumed by the reaction before all the moles of hydroxide anions get the chance to react.

After the neutralization reaction is complete, you will be left with

n_("H"^(+)) = "0.00250 moles" - "0.00250 moles" = "0 moles H"^(+)

n_("OH"^(-)) = "0.00300 moles" - "0.00250 moles" = "0.00050 moles OH"^(-)

The total volume of the solution will be

${V}_{\text{total" = "12.50 mL" + "15.00 mL" = "27.50 mL}}$

The concentration of hydroxide anions in the resulting solution will be

["OH"^(-)] = "0.00050 moles"/(27.5 * 10^(-3)"L") = "0.01818 M"

Now, the pOH of the solution will be

"pOH" = - log(["OH"^(-)])

$\text{pOH} = - \log \left(0.01818\right) = 1.740$

As you know, an aqueous solution at room temperature has

$\textcolor{p u r p \le}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{pH " + " pOH} = 14} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This means that the pH of the solution will be

$\text{pH} = 14 - 1.740 = \textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{12.260} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three decimal places.

Sep 27, 2016

For part (c)

$\textsf{p H = 9.25}$

#### Explanation:

You can see here that we have a mixture of a weak base and its salt. This tells us that we have an alkaline buffer.

Ammonium ions dissociate:

sf(NH_4^(+)rightleftharpoonsNH_3+H^(+)

For which

$\textsf{{K}_{a} = \frac{\left[N {H}_{3}\right] \left[{H}^{+}\right]}{\left[N {H}_{3}\right]} = 5.62 \times {10}^{- 10} \textcolor{w h i t e}{x} \text{mol/l}}$

Please note that these are equilibrium concentrations.

To find the pH we need to find $\textsf{\left[{H}^{+}\right]}$.

Rearranging gives:

$\textsf{\left[{H}^{+}\right] = {K}_{a} \times \frac{\left[N {H}_{3}\right]}{\left[N {H}_{4}^{+}\right]}}$

The initial moles is given by:

$\textsf{{n}_{N {H}_{3}} = c \times v = 0.200 \times \frac{10.00}{1000} = 0.002}$

$\textsf{{n}_{N {H}_{4}^{+}} = c \times v = 0.2 \times \frac{10.00}{1000} = 0.002}$

Now check the value of $\textsf{{K}_{a}}$. It is so small that we can assume that the initial number of moles is a very close approximation to the equilibrium moles.

Because the total volume is common to both $\textsf{N {H}_{3}}$ and $\textsf{N {H}_{4}^{+}}$ solutions we do not need to divide moles by volume to get concentration since the volumes will cancel.

So we can write:

sf([H^+]=5.62xx10^(-10)xx(cancel(0.002))/(cancel(0.002))

$\therefore$$\textsf{p H = - \log \left[{H}^{+}\right] = - \log \left[5.62 \times {10}^{- 10}\right] = 9.25}$

Sep 28, 2016

Here's part $d$. I got a $\text{pH}$ of $4.92$.

By inspection, $\text{NaOH}$ is a strong base, as it dissociates to release ${\text{OH}}^{-}$ ions into solution. It contains 1 ${\text{OH}}^{-}$ that corresponds with one ${\text{H}}^{+}$ in acetic acid.

Both the acid and base have one equivalent of ${\text{H}}^{+}$ or ${\text{OH}}^{-}$ respectively, so there is a 1:1 molar ratio. Thus, the base neutralizes the weak acid ${\text{HC"_2"H"_3"O}}_{2}$ to an equimolar extent.

What we then have to do is find out how much ${\text{H}}^{+}$ gets generated by acetic acid, and neutralize it with the $\text{NaOH}$. With an "ICE Table", we determine the quantity of ${\text{H}}^{+}$ to which base is added later.

${\text{HC"_2"H"_3"O"_2(aq) rightleftharpoons "H"^(+)(aq) + "C"_2"H"_3"O}}_{2}^{-} \left(a q\right)$

$\text{I"" "0.200" "" "" "" "" "0" "" "" "" "" } 0$
$\text{C"" "-x" "" "" "" "" "+x" "" "" "" } + x$
$\text{E"" "0.200-x" "" "" "x" "" "" "" "" } x$

Using the ${K}_{a}$ of acetic acid, which is well-known to be $1.8 \times {10}^{- 5}$, we have the equilibrium expression

${K}_{a} = 1.8 \times {10}^{- 5} = \left(\left[{\text{H"^(+)]["C"_2"H"_3"O"_2^(-)])/(["HC"_2"H"_3"O}}_{2}\right]\right) = \frac{{x}^{2}}{0.200 - x}$

and rearrange to the quadratic equation form

${x}^{2} + 1.8 \times {10}^{- 5} x - 0.200 \cdot 1.8 \times {10}^{- 5} = 0$,

which when solved via the quadratic formula, gives you the two solutions

$x = {\text{0.001888 M H}}^{+}$,
$x = - {\text{0.001906 M H}}^{+}$,

the former of which makes physical sense (you cannot have a negative concentration). So, that is the quantity of ${\text{H}}^{+}$ that would be reacted with $\text{NaOH}$ according to the following chemical reaction:

$\textcolor{g r e e n}{\text{H"^(+)(aq) + "NaOH"(aq) -> "Na"^(+)(aq) + "H"_2"O} \left(l\right)}$

We are using:

• $\text{15.00 mL 0.200 M NaOH}$
• ${\text{25.00 mL 0.200 M HC"_2"H"_3"O}}_{2}$

These are 1:1 acids/bases, so we first realize that we started with

${\text{0.200 mol"/"L" xx "0.01500 L" = "0.00300 mols OH}}^{-}$,

and we have

${\text{0.001888 mols"/"L" xx "0.025 L" = "0.0000472 mols H}}^{+}$.

Furthermore, the weak acid dissociated to form a buffer between ${\text{HC"_2"H"_3"O}}_{2}$ and ${\text{C"_2"H"_3"O}}_{2}^{-}$, which means when we neutralize the acid, we should consider the Henderson-Hasselbalch equation to determine the new $\text{pH}$!

"pH" = -log(K_a) + log\frac(["base"])(["HC"_2"H"_3"O"_2])

The acid we started with has been neutralized by the $\text{NaOH}$, which was the limiting reagent, so there is still ${\text{HC"_2"H"_3"O}}_{2}$ left. This quantity is:

${\text{0.200 mols"/"L" xx "0.025 L" - "0.0000472 mols H}}^{+}$

$=$ $\text{0.00495 mols acid}$

We now determine how much of this acid is also neutralized by the base.

["OH"^(-)] = 0.00300 - "0.0000472 mols OH"^(-) = "0.00295 mols additional base"

So the ${\text{OH}}^{-}$ neutralizes another $\text{0.00295 mols acid}$ to give ${\text{0.00200 mols HC"_2"H"_3"O}}_{2}$ left.

Incorporate that produced acetate into the ${n}_{{\text{C"_2"H"_3"O}}_{2}^{-}}$ to find:

${n}_{\text{base" = "0.00295 mols C"_2"H"_3"O"_2^(-) "generated" + "0.0000472 mols C"_2"H"_3"O"_2^(-) "present" = "0.00300 mols base overall}}$

We have the ions sharing the same solution, so we can choose to use the $\text{mol}$s instead of the concentrations. From the HH equation:

$\textcolor{b l u e}{\text{pH") = -log(1.8xx10^(-5)) + log\frac("0.00300 mols")("0.00495 mols" - "0.00295 mols}} \approx \textcolor{b l u e}{4.92}$

Sep 28, 2016

For Question (f)
$p H = 11.01$

#### Explanation:

Question

To calculate the pH of the resulting solution when 30mL0.1M KOH soution is mixed with 15 mL 0.2M HCN solution

30mL0.1M KOH $\equiv \left(30 \times 0.1\right) = 3 m m o l K O H$

15mL0.2M HCN $\equiv \left(15 \times 0.2\right) = 3 m m o l H C N$

Now the equation of the reaction is

$K O H + H C N \to K C N + {H}_{2} O$

This equation reveals that 1mol KOH netralises 1 mol HCN and forms 1 mol KCN.So in our given mixture 3 mmol KCN will be formed.There will be no excess acid or base. The volume of the resulting mixture will be $\left(30 + 15\right) m L = 45 m L$

Hence the concentration of resulting KCN solution
$c = \frac{3 m m o l}{45 m L} = \frac{3}{45} M = \frac{1}{15} M$

The KCN being a salt of weak acid HCN and strong base KOH,it will be hydrolysed and the pH of this solution is related with its concentration (c) as follows.

pH=7+1/2pK_a+1/2logc ......color(red)((1))
where ${K}_{a}$ is the acid dissociation constant of the weak acid HCN and the value of $p {K}_{a} = 9.2$

So inserting these values in the above equation we get

$p H = 7 + \frac{1}{2} \times 9.2 + \frac{1}{2} \log \left(\frac{1}{15}\right)$

$= 7 + 4.6 - 0.5 \times \log 15$

$= 11.6 - 0.5 \times 1.17 \approx 11.01$

Deduction of relation (1)

Anionic hydrolysis of strong conjugate base $C {N}^{-}$ with its ICE-table

$C {N}^{\text{-" +H_2O stackrel(K_h) (rightleftharpoons)HCN" "+" }} O {H}^{-}$

$I \text{ "" "c" "" "" "" "" "" "" " 0" "" "" "" "" } 0$

$C \text{ "-ch" "" "" "" "" "" " ch" "" "" "" } c h$

$E \text{ "c-ch" "" "" "" "" "" " ch" "" "" "" } c h$

Where c is concentration of $C {N}^{-}$ before hydrolysis and h is the degree of hydrolysis.

So ${K}_{h} = \frac{\left[H C N\right] \left[O H\right]}{\left[C {N}^{-}\right]} \ldots \ldots \textcolor{red}{\left(2\right)}$

$\implies {K}_{h} = \frac{\left(c h\right) \left(c h\right)}{c \left(1 - h\right)} \approx c {h}^{2}$

$\implies h = \sqrt{{K}_{h} / c} \ldots \ldots \textcolor{red}{\left(3\right)}$

Considering ionic equilibrium of dissociation of $H C N$ in aqueous solution.

$H C N + {H}_{2} O \stackrel{{K}_{a}}{r} i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + C {N}^{-}$

K_a = ([H_3O^(+) ][CN^-])/([HCN]) ......color(red)((4))
Considering ionic equilibrium of dissociation of ${H}_{2} O$

${H}_{2} O + {H}_{2} O \stackrel{{K}_{w}}{r} i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + O {H}^{-}$

The ionic product of water
K_w=([H_3O^+[OH^-])=10^-14" ".....color(red)((5))
Taking log

$- \log {K}_{w} = - \log \left({10}^{-} 14\right) = 14$

$\implies p {K}_{w} = - \log \left(\left[{H}_{3} {O}^{+}\right] \left[O {H}^{-}\right]\right) = \log \left({10}^{-} 14\right) = 14$

$\implies p {K}_{w} = p H + p O H = - \log \left({10}^{-} 14\right) = 14 \ldots \ldots \textcolor{red}{\left(6\right)}$

Now by (4) and (5)

${K}_{w} / {K}_{a} = \frac{\left[H C N\right] \left[O H\right]}{\left[C {N}^{-}\right]} = {K}_{h}$

Again

$\left[O {H}^{-}\right] = c h = c \times \sqrt{{K}_{h} / c} = \sqrt{{K}_{h} c}$

$p O H = - \log \left[O {H}^{-}\right] = - \log \sqrt{{K}_{h} c}$

$\implies p O H = - \log \sqrt{{K}_{h} c}$

$= - \frac{1}{2} \log {K}_{h} - \frac{1}{2} \log c$

$\implies p O H = \frac{1}{2} \left(- \log \left({K}_{w} / {K}_{a}\right) - \log c\right)$

$\implies p O H = \frac{1}{2} \left(- \log {K}_{w} + \log {K}_{a} - \log c\right)$

$\implies p O H = \frac{1}{2} \left(p {K}_{w} - p {K}_{a} - \log c\right)$

$\implies 14 - p H = \frac{1}{2} \left(14 - p {K}_{a} - \log c\right)$

$\implies 14 - p H = 7 - \frac{1}{2} \left(p {K}_{a} + \log c\right)$

$\implies p H = 7 + \frac{1}{2} \left(p {K}_{a} + \log c\right)$

Sep 28, 2016

f) pH = 2.04

#### Explanation:

We are adding a strong acid to a weak base, so our first task is to calculate the moles of reactants and of products.

The reaction will go to completion until one of the reactants is used up.

The relation is

color(blue)(bar(ul(|color(white)(a/a) "moles" = "litres × molarity"color(white)(a/a)|)))" "

Let's use an ICE table to keep track of the calculations.

$\textcolor{w h i t e}{m m m m m m} \text{HCl"color(white)(m) +color(white)(m) "NH"_3color(white)(m) →color(white)(mll) "NH"_4^+color(white)(m) + color(white)(ml)"Cl"^"-}$
$\text{I/mol:"color(white)(mll)"0.003 000"color(white)(m)"0.002 500} \textcolor{w h i t e}{m m m m l} 0 \textcolor{w h i t e}{m m m m m l} 0$
$\text{C/mol:"color(white)(m)"-0.002 500"color(white)(ll)"-0.002 500"color(white)(mll)"+0.002 500"color(white)(ll)"+0.002 500}$
$\text{E/mol:" color(white)(ml)"0.000 500"color(white)(mmm)0color(white)(mmmmll)"0.002 500"color(white)(mll)"0.002 500}$

We end up with a solution of 0.000 500 mol $\text{HCl}$ and 0.002 500 mol ${\text{NH}}_{4}^{+}$ in 55.00 mL of solution.

The strong acid $\text{HCl}$ will suppress and overwhelm the dissociation of the weak acid ${\text{NH}}_{4}^{+}$.

We have effectively a dilute solution of $\text{HCl}$, which ionizes completely in solution.

$\text{HCl" color(white)(l)+ "H"_2"O" → "H"_3"O"^+ + "Cl"^"-}$

["H"_3"O"^+] = "0.000 500 mol"/"0.055 00 L" ="0.009 091 mol/L"

"pH" = "-"log["H"_3"O"^+] = "-"log("0.009 091") = 2.04

Sep 30, 2016

For part b), I got pH = 9.72.

#### Explanation:

Our first step is to calculate the molarity of the solution.

${\text{Moles of Na"_2"SO"_3 = 3.50 color(red)(cancel(color(black)("g Na"_2"SO"_3))) × ("1 mol Na"_2"SO"_3)/(126.04 color(red)(cancel(color(black)("g Na"_2"SO"_3)))) = "0.027 77 mol Na"_2"SO}}_{3}$

$\text{Molarity" = "moles"/"litres" = ("0.027 77 mol Na"_2"SO"_3)/"1.50 L" = "0.018 51 mol/L}$

${\text{Na"_2"SO}}_{3}$ is the salt of the weak acid ${\text{H"_2"SO}}_{3}$.

$\text{H"_2"SO"_3 + "H"_2"O" ⇌ "H"_3"O"^+ + "HSO"_3^"-"; K_text(a1) = 1.2 × 10^"-2}$

$\text{HSO"_3^"-" + "H"_2"O" ⇌ "H"_3"O"^+ + "SO"_3^"2-"; K_text(a2) = 6.6 × 10^"-8}$

We have a solution of $\text{SO"_3^"2-}$, so we use the second equilibrium.

$\text{HSO"_3^"-}$ is a weak acid, and $\text{SO"_3^"2-}$ is its conjugate base.

$\text{SO"_3^"2-" + "H"_2"O" ⇌ "HSO"_3^"-" + "OH"^-}$

${K}_{\text{b" = K_"w"/K_"a2" = (1.00 × 10^"-14")/(6.6 × 10^"-8") = 1.52 × 10^"-7}}$

Now, we can set up an ICE table.

$\textcolor{w h i t e}{m m m m m m m m} \text{SO"_3^"2-" + "H"_2"O" ⇌ "HSO"_3^"-" + "OH"^-}$
$\text{I/mol·L"^"-1":color(white)(mll)"0.018 51} \textcolor{w h i t e}{m m m m m m l} 0 \textcolor{w h i t e}{m m m} 0$
$\text{C/mol·L"^"-1":color(white)(mmm)"-"xcolor(white)(mmmmmmm)"+"xcolor(white)(mm)"+} x$
$\text{E/mol·L"^"-1":color(white)(ll)"0.018 51-} x \textcolor{w h i t e}{m m m m m m} x \textcolor{w h i t e}{m m m} x$

${K}_{\text{b" = (["HSO"_3^"-"]["OH"^"-"])/(["SO"_3^"2-"]) = (x·x)/("0.018 51-"x) = x^2/("0.018 51-"x) = 1.52 × 10^"-7}}$

A common rule of thumb is that we can ignore $x$ if the initial concentration is greater than 400 × K_"b".

"0.018 51"/(1.52 × 10^"-7") = 1.22 × 10^5, so x ≪ "0.018 51"

${x}^{2} / \text{0.018 51" = 1.52 × 10^"-7}$

${x}^{2} = \text{0.018 51" × 1.52 × 10^"-7" = 2.81 × 10^"-9}$

x = 5.30 × 10^"-5"

["OH"^"-"] = 5.30 × 10^"-5"color(white)(l) "mol/L"

"pOH" = "-log"["OH"^"-"] = "-log"(5.30 × 10^"-5") = 4.28

$\text{pH" = "14.00 - pOH" = "14.00 - 4.28} = 9.72$