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
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
6 Answers
For part (a):
Explanation:
I'm going to start this off by solving part (a).
In this case, you're mixing hydrochloric acid,
You are told that all the base dissolves, which means that the solution contains twice as many moles of hydroxide anions,
#"Ca"("OH")_ (color(red)(2)(s)) -> "Ca"_ ((aq))^(2+) + color(red)(2)"OH"_ ((aq))^(-)#
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"HCl"_ ((aq)) + "Ca"("OH")_ (2(s)) -> "CaCl"_ (2(aq)) + 2"H"_ 2"O"_ ((l))#
Because you assume that all the base is dissolved, you can write the net ionic equation like this
#2"H"_ ((aq))^(+) + 2"OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))#
which, of course, gets you
#"H"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "H"_ 2"O"_ ((l))#
So, the reaction consumes equal numbers of moles of hydrogen ions,
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_"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"^(-)])#
#"pOH" = - log(0.01818) = 1.740#
As you know, an aqueous solution at room temperature has
#color(purple)(bar(ul(|color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|)))#
This means that the pH of the solution will be
#"pH" = 14 - 1.740 = color(green)(bar(ul(|color(white)(a/a)color(black)(12.260)color(white)(a/a)|)))#
The answer is rounded to three decimal places.
For part (c)
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:
For which
Please note that these are equilibrium concentrations.
To find the pH we need to find
Rearranging gives:
The initial moles is given by:
Now check the value of
Because the total volume is common to both
So we can write:
Here's part
By inspection,
Both the acid and base have one equivalent of
What we then have to do is find out how much
#"HC"_2"H"_3"O"_2(aq) rightleftharpoons "H"^(+)(aq) + "C"_2"H"_3"O"_2^(-)(aq)#
#"I"" "0.200" "" "" "" "" "0" "" "" "" "" "0#
#"C"" "-x" "" "" "" "" "+x" "" "" "" "+x#
#"E"" "0.200-x" "" "" "x" "" "" "" "" "x#
Using the
#K_a = 1.8xx10^(-5) = (["H"^(+)]["C"_2"H"_3"O"_2^(-)])/(["HC"_2"H"_3"O"_2]) = (x^2)/(0.200 - x)#
and rearrange to the quadratic equation form
#x^2 + 1.8xx10^(-5)x - 0.200*1.8xx10^(-5) = 0# ,
which when solved via the quadratic formula, gives you the two solutions
#x = "0.001888 M H"^(+)# ,
#x = -"0.001906 M H"^(+)# ,
the former of which makes physical sense (you cannot have a negative concentration). So, that is the quantity of
#color(green)("H"^(+)(aq) + "NaOH"(aq) -> "Na"^(+)(aq) + "H"_2"O"(l))#
We are using:
#"15.00 mL 0.200 M NaOH"# #"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
#"0.200 mol"/"L" xx "0.01500 L" = "0.00300 mols OH"^(-)# ,
and we have
#"0.001888 mols"/"L" xx "0.025 L" = "0.0000472 mols H"^(+)# .
Furthermore, the weak acid dissociated to form a buffer between
#"pH" = -log(K_a) + log\frac(["base"])(["HC"_2"H"_3"O"_2])#
The acid we started with has been neutralized by the
#"0.200 mols"/"L" xx "0.025 L" - "0.0000472 mols H"^(+)#
#=# #"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
Incorporate that produced acetate into the
#n_"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
#color(blue)("pH") = -log(1.8xx10^(-5)) + log\frac("0.00300 mols")("0.00495 mols" - "0.00295 mols") ~~ color(blue)(4.92)#
For Question (f)
Explanation:
Question
To calculate the pH of the resulting solution when 30mL0.1M KOH soution is mixed with 15 mL 0.2M HCN solution
Answer
30mL0.1M KOH
15mL0.2M HCN
Now the equation of the reaction is
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
Hence the concentration of resulting KCN solution
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.
where
So inserting these values in the above equation we get
#=7+4.6-0.5xxlog15#
#=11.6-0.5xx1.17~~11.01#
Deduction of relation (1)
Anionic hydrolysis of strong conjugate base
#CN^"-" +H_2O stackrel(K_h) (rightleftharpoons)HCN" "+" "OH^-#
Where c is concentration of
So
Considering ionic equilibrium of dissociation of
Considering ionic equilibrium of dissociation of
The ionic product of water
Taking log
Now by (4) and (5)
Again
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.
We end up with a solution of 0.000 500 mol
The strong acid
We have effectively a dilute solution of
For part b), I got pH = 9.72.
Explanation:
Our first step is to calculate the molarity of the solution.
We have a solution of
Now, we can set up an ICE table.
A common rule of thumb is that we can ignore
∴
∴