Question

Question: At^{25$°$C, 50.00mL}a ^0.1000Msolution of maleic acid, a diprotic acid whose ionization constants are ^K_a1=1.42$\times$10^-2and^K_a2=8.57$\times$10^-7, is titrated with a ^{0.1000M NaOH}solution. Calculate the following volumes of added base:

^{0, 5.00, 25.00, 50.00, 75.00, 99.90, 100.00}and105.00 mL.

Short answer

p H at the following volumes of the added base are:

p H =1.51

pH =1.61

pH =1.85

p H=3.96

p H=6.07

p H=8.76

p H=9.3

p H=11.51

Step-by-step solution

The first and second reactions.

The first reaction:

H2A+H2O$\to$HA^-+H3O⁺

The ^$K_{a_1}$reaction is 1.42$\times$10^-2.

The second reaction:

HA^-+H2O$\to$A^2-+H3O⁺

The _{$K_{a_2}$} reaction is8.57$\times$10^-7.

First and second equivalency point.

First, calculate $\mathrm{\eta }$ of $H_{2}A$:

n( H_-2 A )={HA}$\times$V

n( H_-2 A )=0.1M$\times$50$\times$10^-3L

n( H_-2 A )=5$\times$10^-3 mol

The volume of the firstequivalencepoint must be calculated. The quantity NaOH is exactly enough to devour the at the ^HA equivalence point. So, here V_e:

_.

The second equivalent point is:

_{$V_{e_1}$}=$\frac{1}{2}\times V_{e_2}$

$V_{e_2}=2\times V_{e_1}$

$V_{e_2}$= 2$\times$50 mL

$V_{e_2}=100mL$

Equal concentration.

$\times$When V(NaOH)=0

H_-2A$\to$HA^-+HA^-+H⁺

If put^x,

_{$K_{a_1}$= $\frac{x^2}{[H2A]-x}$}

$1.42\times {10}^{-2}$= $\frac{x^2}{0.1-x}$

x²+1.42$\times$10^-2x-1.42$\times$10^-3=0

x=0.0312

Then it acts as an equal concentration.

p H is calculated

p H= -log([H⁺])

p H= -log(0.0312)

pH= 1.51

V(NaOH)=5 m L , then n of OH will be

n (OH^-) ={NaOH}$\times$V(NaOH)

= 0. 1M $\times$5 $\times$10^-3 L

n (OH^-) = 5$\times$10^-4mol$\times$

Reaction will be H2A+OH $\to$HA +H2O

The ^{$\mathrm{\eta }$}of H²A and HA^- is

n(H.2 A)=n(H.2 A) _initial-n (OH^.)

n(H.2 A)= (5$\times$10^-3-5$\times$10^-4) mol

n(H.2 A)= 4.5$\times$10^-3mol

n (HA-) =n (HA-) _INITIAL+n(OH-)

n (HA-) = 0+5.10^-4 mol

n (HA-) = 5.10^-4mol$\times$

The initial concentration

[H.AA]= $\frac{4.5\times {10}^{-3}\mathrm{mol}}{55\times {10}^{-3}\mathrm{L}}$

[H.2 A] = 0.0818M

[HA-] =$\frac{5\times {10}^{-4}mol}{55\times {10}^{-3}L}$

[HA.] = 9.09 $\times$10^-3M

localid="1664028675749" $\begin{array}{c}{\text{K}}_{{\text{a}}_{\text{1}}}\text{=}\frac{\text{x×}\left(\left[{\text{HA}}^{\text{-}}\right]\text{+x}\right)}{\left[{\text{H}}_{\text{-}}\text{2A}\right]\text{-x}}\\ {\text{1.42×10}}^{\text{-2}}\text{=}\frac{\text{x×}\left({\text{9.09×10}}^{\text{-3}}\text{M+x}\right)}{\text{0.0818M-x}}\\ {\text{1.162×10}}^{\text{-3}}{\text{-1.42×10}}^{\text{-2}}{\text{x=9.09×10}}^{\text{-3}}{\text{x+x}}^{\text{2}}\\ {\text{x}}^{\text{2}}{\text{+0.02329x-1.162×10}}^{\text{-3}}\text{=0x}\\ \text{=0.0244M=}\left[{\text{H}}^{\text{+}}\right]\end{array}$

$\begin{array}{c}\text{pH=-log}\left(\left[{\text{H}}^{\text{+}}\right]\right)\\ \text{pH=-log(0.0244)}\\ \text{pH=1.61}\end{array}$

Concentrations are equal.

When$\text{V(NaOH)=25ml}$

$\left[{\text{H}}_{\text{-}}\text{2A}\right]\text{=}\left[{\text{HA}}^{\text{-}}\right]$

When${\text{pH=pK}}_{{\text{a}}_{\text{1}}}{\text{:pK}}_{{\text{a}}_{\text{1}}}$

$\begin{array}{c}{\text{pK}}_{{\text{a}}_{\text{1}}}\text{=-log}\left({\text{K}}_{{\text{a}}_{\text{1}}}\right)\\ {\text{pK}}_{{\text{a}}_{\text{1}}}\text{=-log}\left({\text{1.42×10}}^{\text{-2}}\right)\\ {\text{pK}}_{{\text{a}}_{\text{1}}}\text{=1.85pH=1.85}\end{array}$

If$\text{V(NaOH)=50mL}$

$\text{pH=}\frac{\text{1}}{\text{2}}\text{×}\left({\text{pK}}_{{\text{a}}_{\text{1}}}{\text{+pK}}_{{\text{a}}_{\text{2}}}\right)$

${\text{pK}}_{{\text{a}}_{\text{2}}}$ will be

$\begin{array}{c}{\text{pK}}_{{\text{a}}_{\text{2}}}\text{=-log}\left({\text{K}}_{{\text{a}}_{\text{2}}}\right)\\ {\text{pK}}_{{\text{a}}_{\text{2}}}\text{=-log}\left({\text{8.57×10}}^{\text{-7}}\right)\\ {\text{pK}}_{{\text{a}}_{\text{2}}}\text{=6.07}\end{array}$

$\begin{array}{c}pH=\frac{1}{2}\cdot (1.85+6.07)\\ pH=3.96\end{array}$

The concentrations are:

$\begin{array}{c}\left[{\text{HA}}^{\text{-}}\right]\text{=}\frac{\text{n}{\left({\text{HA}}^{\text{-}}\right)}_{\text{initial}}\text{-n}\left({\text{OH}}^{\text{-}}\right)}{{\text{V}}_{\text{total}}}\\ \left[{\text{HA}}^{\text{-}}\right]\text{=}\frac{{\text{5×10}}^{\text{-3}}{\text{mol-2.5×10}}^{\text{-3}}\text{mol}}{\text{0.125L}}\\ \left[{\text{A}}^{\text{2-}}\right]\text{=}\frac{\text{n}{\left({\text{A}}^{\text{2-}}\right)}_{\text{initial}}\text{+n}\left({\text{OH}}^{\text{-}}\right)}{{\text{V}}_{\text{total}}}\\ \left[{\text{A}}^{\text{2-}}\right]\text{=}\frac{{\text{0+2.5×10}}^{\text{-3}}\text{mol}}{\text{0.125L}}\\ \left[{\text{A}}^{\text{2-}}\right]\text{=0.02M}\end{array}$

The concentrations are equal:

$\left[{\text{HA}}^{\text{-}}\right]\text{=}\left[{\text{A}}^{\text{-}}\right]$

$\text{pH}$is equal to${\text{pK}}_{{\text{a}}_{\text{2}}}$

$\begin{array}{c}{\text{pH=pK}}_{{\text{a}}_{\text{2}}}\\ \text{=6.07}\end{array}$

When$\text{V}\left(\text{NaOH}\right)=99.90$,$\text{49.90mL}$ will be as the first equivalence point.

$\begin{array}{c}\text{n}\left({\text{OH}}^{\text{-}}\right)\text{=[NaOH]×V(NaOH)}\\ {\text{=0.1M×49.90×10}}^{\text{-3}}\text{L}\\ \text{n}\left({\text{OH}}^{\text{-}}\right){\text{=4.99×10}}^{\text{-3}}\text{mol}\end{array}$

${\text{HA}}^{\text{-}}{\text{+OH}}^{\text{-}}\to {\text{A}}^{\text{2-}}{\text{+H}}_{\text{-}}\text{2O}$

$\text{n}$ of ${\text{HA}}^{\text{-}}$will be

$\text{n}\left({\text{HA}}^{\text{-}}\right)\text{=n}{\left({\text{HA}}^{\text{-}}\right)}_{\text{initiol}}\text{-n}\left({\text{OH}}^{\text{-}}\right)$

The initial concentrations.

$\begin{array}{c}\text{n}\left({\text{A}}^{\text{2-}}\right)\text{=n}{\left({\text{A}}^{\text{2-}}\right)}_{\text{initial}}\text{+n}\left({\text{OH}}^{\text{-}}\right)\\ \text{n}\left({\text{A}}^{\text{2-}}\right){\text{=0+4.99×10}}^{\text{-3}}\text{mol}\\ \text{n}\left({\text{A}}^{\text{2-}}\right){\text{=4.99×10}}^{\text{-3}}\text{mol}\end{array}$

The initial concentrations are:

$\begin{array}{c}\left[{\text{HA}}^{\text{-}}\right]\text{=}\frac{{\text{1×10}}^{\text{-5}}\text{mol}}{{\text{149.90×10}}^{\text{-3}}\text{L}}\\ \left[{\text{HA}}^{\text{-}}\right]{\text{=6.67×10}}^{\text{-5}}\text{M}\\ \left[{\text{A}}^{\text{2-}}\right]\text{=}\frac{{\text{4.99×10}}^{\text{-3}}\text{mol}}{{\text{149.90×10}}^{\text{-3}}\text{L}}\\ \left[{\text{A}}^{\text{2-}}\right]\text{=0.0333M}\end{array}$

${\text{K}}_{{\text{a}}_{\text{2}}}\text{=}\frac{\text{x×}\left(\left[{\text{A}}^{\text{2-}}\right]\text{+x}\right)}{\left[{\text{HA}}^{\text{-}}\right]\text{-x}}$

$\begin{array}{c}{\text{8.57×10}}^{\text{-7}}\text{=}\frac{\text{x×(0.0333+x)}}{{\text{6.67×10}}^{\text{-5}}\text{-x}}\\ {\text{5.716×10}}^{\text{-11}}{\text{-8.57×10}}^{\text{-7}}{\text{x=0.0333x+x}}^{\text{2}}\\ {\text{x}}^{\text{2}}{\text{+0.0333x-5.716×10}}^{\text{-11}}\text{=0}\\ {\text{x=1.72×10}}^{\text{-9}}\text{M}\\ \text{=}\left[{\text{H}}^{\text{+}}\right]\\ \text{pH=-log}\left(\left[{\text{H}}^{\text{+}}\right]\right)\\ \text{pH=-log}\left({\text{1.72×10}}^{\text{-9}}\right)\\ \text{pH=8.76}\end{array}$

When $\text{V(NaOH)=100mL}$ the concentrations are

${\text{A}}^{\text{2-}}{\text{+H}}_{\text{-}}\text{2O}\to {\text{HA}}^{\text{-}}{\text{+OH}}^{\text{-}}$

${\text{K}}_{\text{b}}$ concentration is

$\begin{array}{c}{\text{K}}_{\text{b}}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{{\text{a}}_{\text{2}}}}\text{=}\frac{{\text{1×10}}^{\text{-14}}}{{\text{8.57×10}}^{\text{-7}}}\\ {\text{K}}_{\text{b}}{\text{=1.17×10}}^{\text{-8}}\text{n}\end{array}$

$\begin{array}{c}{\text{K}}_{\text{b}}\text{=}\frac{\left[{\text{HA}}^{\text{-}}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[{\text{A}}^{\text{2-}}\right]}\\ {\text{K}}_{\text{b}}\text{=}\frac{\text{x×x}}{\text{0.0333M-x}}\\ {\text{1.17×10}}^{\text{-8}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0.0333-x}}\\ {\text{x}}^{\text{2}}{\text{+1.17×10}}^{\text{-8}}{\text{x-3.896×10}}^{\text{-10}}\text{=0}\\ {\text{x=1.97×10}}^{\text{-5}}\text{M}\\ \text{=}\left[{\text{OH}}^{\text{-}}\right]\end{array}$

Determination PH .

$\begin{array}{c}\text{pOH=-log}\left(\left[{\text{OH}}^{\text{-}}\right]\right)\\ \text{pOH=-log}\left({\text{1.97×10}}^{\text{-5}}\right)\\ \text{pOH=4.7}\end{array}$

$\begin{array}{c}\text{pH=14-pOH}\\ \text{=14-4.7}\\ \text{pH=9.3}\end{array}$

When,$\text{V(NaOH)=105mL}$

$\begin{array}{c}\left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{\left({\text{V(added)-V}}_{{\text{e}}_{\text{2}}}\right)\text{×[NaOH]}}{\text{V(total)}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{\left({\text{105×10}}^{\text{-3}}{\text{-100×10}}^{\text{-3}}\right)\text{L×0.1M}}{{\text{155×10}}^{\text{-3}}\text{L}}\\ \left[{\text{OH}}^{\text{-}}\right]{\text{=3.23×10}}^{\text{-3}}\text{M}\end{array}$

Then the $\text{pOH}$value is,

$\begin{array}{c}\text{pOH=-log}\left(\left[{\text{OH}}^{\text{-}}\right]\right)\\ \text{pOH=-log}\left({\text{3.23×10}}^{\text{-3}}\right)\\ \text{pOH=2.49}\end{array}$

Then$\text{pH}$ is reacted as,

$\begin{array}{c}\text{pH=14-pOH}\\ \text{=14-2.49}\\ \text{pH=11.51}\end{array}$

Therefore, the result is determined as,

$\begin{array}{c}\text{pH=1.51}\\ \text{pH=1.61}\\ \text{pH=1.85}\\ \text{pH=3.96}\\ \text{pH=6.07}\\ \text{pH=8.76}\\ \text{pH=9.3}\\ \text{pH=11.51}\end{array}$