# Question 07316

Sep 13, 2015

$p H = 0.35$

#### Explanation:

First we find how many moles of $H C l$ we have initially.

$c = \frac{n}{v}$

So: $n = c \times v$

So initial moles $H C l = 25.0 \times 0.723 = 18.075 \text{mmol}$

(its mmol since there are 1000ml in 1L)

The alkali neutralises the acid:

$H C {l}_{\left(a q\right)} + K O {H}_{\left(a q\right)} \rightarrow K C {l}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)}$

This tells use that for every mole of $K O H$ added, 1 mole of $H C l$ is neutralised.

The no. moles $K O H$ added $= 10 \times 0.273 = 2.73 \text{mmol}$

So the no. moles $H C l$ remaining must be:

$18.075 - 2.73 = 15.34 \text{mmol"=15.34xx10^(-3)"mol}$

The new volume $= 25.0 + 10.0 = 35.0 \text{ml"=35xx10^(-3)"L}$

$c = \frac{n}{v}$

So the concentration of $H C l$ is now:

$\frac{15.34 \times \cancel{{10}^{- 3}}}{35 \times \cancel{{10}^{- 3}}}$

$= 0.44 \text{mol/L}$

$p H = - \log \left[{H}_{\left(a q\right)}^{+}\right]$

So:

$p H = - \log \left[0.44\right]$

$p H = 0.35$

Sep 13, 2015

The pH will be $0.358$.

#### Explanation:

You're titrating hydrochloric acid, $\text{HCl}$, a strong acid, with potassium hydroxide, $\text{KOH}$, a strong base, which means that the pH of the resulting solution will be equal to $7$ at the equivalence point.

So right from the start, you know that if the neutralization is complete, the pH of the solution will be equal to $7$.

But is the equivalence point reached or not?

The balanced chemical equation for this neutralization reaction is

${\text{HCl"_text((aq]) + "KOH"_text((aq]) -> "KCl"_text((aq]) + "H"_2"O}}_{\textrm{\left(l\right]}}$

The $1 : 1$ mole ratio that exists between hydrochloric acid and potassium chloride tells you that the equivalence point is reached when the number of moles of the acid is equal to the number of moles of the base.

If you have fewer moles of base, the solution will remain acidic, which means that you can expect to have $\text{pH < 7}$.

If you have more moles of base, the solution will become basic, which corresponds to $\text{pH > 7}$.

Use the molarities and volumes of the two solution to find the number of moles of each

$n = \frac{C}{V} \implies n = C \cdot V$

${n}_{H C l} = \text{0.723 M" * 25.0 * 10^(-3)"L" = "0.01808 moles HCl}$

and

${n}_{K O H} = \text{0.273 M" * 10.0 * 10^(-3)"L" = "0.002730 moles KOH}$

This means that the reaction will completely consume the added potassium hydroxide, leaving you with

${n}_{H C l} = 0.01808 - 0.002730 = \text{0.01535 moles HCl}$

Since hydrochloric acid is a strong acid, it dissociates completely in aqueous solution to produce hydrogen ions, ${\text{H}}^{+}$, and chloride ions.

This means that the number of moles of hydrogen ions you have in solution will be equal to the number of moles of hydrochloric acid ($1 : 1$ mole ratio again).

The total volume of the solution will be

${V}_{\text{total}} = {v}_{H C l} + {V}_{K O H}$

${V}_{\text{total" = 25.0 + 10.0 = "35.0 mL}}$

The molarity of the hydrogen ions will be

["H"^(+)] = "0.01535 moles"/(35.0 * 10^(-3)"L") = "0.4386 M"

The pH of the solution will thus be

pH_"sol" = -log(["H"^(+)]) = - log(0.4386) = color(green)(0.358)#