# What is the molar concentration of hydrogen ions if 12 drops of a benzoic acid HC_7H_5O_2 solution are titrated with 26 drops of a .40 M KOH solution?

May 31, 2016

["H"^+] = 1.3 × 10^"-9" color(white)(l)"mol/L"

#### Explanation:

I assume that you have titrated to the end point.

The equation for the reaction is

$\text{HA" + "KOH" → "KA" +"H"_2"O}$

Since $\text{HA}$ and $\text{KOH}$ react in a 1:1 molar ratio, we can write

${c}_{a} {V}_{a} = {c}_{b} {V}_{b}$

c_a = c_b × V_b/V_a = "0.40 mol/L" × (26 color(red)(cancel(color(black)("drops"))))/(12 color(red)(cancel(color(black)("drops")))) = "0.867 mol/L"

At neutralization, you have only $\text{KA}$ present.

If there had been no change in volume, its concentration would have been 0.867 mol/L.

However, the volume increased during the titration to $\text{(12 drops + 26 drops) = 38 drops}$.

To calculate the new concentration, we can use the dilution formula

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {c}_{1} {V}_{1} = {c}_{2} {V}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

${c}_{1} = \text{0.867 mol/L"; V_1 = "12 drops}$
c_2 = ?; color(white)(mmmmml)V_2 = "38 drops"

c_2 = c_1 × V_1/V_2 = "0.867 mol/L" × (12 color(red)(cancel(color(black)("drops"))))/(28color(red)(cancel(color(black)( "drops")))) = "0.372 mol/L"

∴ At neutralization, we have a 0.372 mol/L solution of $\text{KA}$.

Calculate the concentration of $\left[{\text{H}}^{+}\right]$ in a 0.372 mol/L solution of KA

$\textcolor{w h i t e}{m m m m m m m m l l} \text{A"^"-" + "H"_2"O" ⇌ "HA" + "OH"^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m m l} 0.372 \textcolor{w h i t e}{m m m m m l l} 0 \textcolor{w h i t e}{m m l l} 0$
$\text{C/mol·L"^"-1":color(white)(mmm) "-} x \textcolor{w h i t e}{m m m m m l} + x \textcolor{w h i t e}{m} + x$
$\text{E/mol·L"^"-1} : \textcolor{w h i t e}{m} 0.372 - x \textcolor{w h i t e}{m m m m l l} x \textcolor{w h i t e}{m m l l} x$

K_b = K_w/K_a = (1.00 × 10^"-14")/(6.46 × 10^"-5") = 1.55 × 10^"-10"

${K}_{b} = \left(\left[\text{HA"]["OH"^"-"])/(["A"^"-}\right]\right)$

1.55 × 10^"-10" = (x × x)/(0.372 -x) - x^2/(0.372-x)

Check if $x$ is negligible

0.372/(1.55 × 10^"-10") = 2.11 × 10^9 > 400; ∴ x « 0.372

1.55 × 10^"-10" = x^2/0.372

x^2 = 0.372 × 1.55 × 10^"-10" = 5.77 × 10^"-11"

x = ["OH"^"-"] = 7.59 ×10^"-6"

["H"^+] = K_w/["OH"^"-"] = (1.00 × 10^"-14")/(7.59 ×10^"-6") = 1.3 × 10^"-9"color(white)(l) "mol/L"