Question

Calculate the \(pH\) at the following points in a titration of \(40\;mL(0.040\;L)\) \(of 0.100 M\) barbituric acid (\({K_a} = 9.8\)\( \times 1{0^{ - 5}}\) ) with\(0.100MKOH\) .

(a) no \(KOH\) added

(b) \(20\;mL of KOH\) solution added

(c) \(39\;mL\) of \(KOH\) solution added

(d) \(40\;mL\) of \(KOH\) solution added

(e) \(41\;mL\) of \(KOH\) solution added

Short answer

a) 2.51

b) 4.01

c) 5.60

d) 8.35

e) 11.09

Step-by-step solution

To find the pH value in no KOH added:

- We have\(0.040\;L\)of\(0.100M{C_4}{H_4}\;{N_2}{O_3}\)

-\({K_a}\)of\({C_4}{H_4}\;{N_2}{O_3}\)\(is\)\({K_a}\)= 9.8 .\(1{0^{ - 5}}\)

(a)Let us calculate the\(pH\)of a given solution

	\(|{C_4}{H_{{4_{}}}}{N_2}{O_3}(aq) \to \)	\({H_3}{O^ + }\left( {aq} \right)\)	\({C_4}{H_4}{N_2}{O_3}^{ - )}(aq)\)
Initial (M)	0.100	0	0
Change (M)	-X	+X	+X
Equilibrium (M)	0.100-X	X	X

\(\begin{array}{l}{K_a} = \frac{{\left[ {1{I_3}{O^ + }} \right] \times \left[ {{C_4}{\mu _3}\;{N_2}O_3^ - } \right]}}{{\left[ {{C_4}{H_4}\;{N_2}{O_3}} \right]}}\\9.8 \times 1{0^{ - 5}} = \frac{{x \times x}}{{0.100 - x}}\\{x^2} = 9.8 \times 1{0^{ - 6}} - 9.8 \times 1{0^{ - 5}}x\\0 = {x^2} + 9.8 \times 1{0^{ - 5}}x - 9.8 \times 1{0^{ - 6}}\\x = 0.00308M\\\left[ {{H_3}{O^ + }} \right] = 0.00308M\end{array}\)

\(Therefore,thepH valueis\)

\(\begin{array}{c}pH = - log\left[ {{H_3}{O^ + }} \right]\\ = - log[0.00308]\\ = 2.51\end{array}\)

To find the pH value in 20ml of KOH solution added:

- We have\(0.040\;L\)of\(0.100M{C_4}{H_4}\;{N_2}{O_3}\)

- and\(20\;mL\)of\(0.100MKOH\)the solution is added

- So, the new volume of a mixture is\(0.060\;L(60\;mL)\)

Let us calculate the\(pH\)value

The concentrations of\({C_4}{H_4}\;{N_2}{O_3}\)and\(KOH\)in a mixture is

\(\begin{array}{l}\left[ {{C_4}{H_4}{N_2}{O_3}} \right] = \frac{{0.040L \times 0.100M}}{{0.060L}} = 0.06667M\\(KOH] = \frac{{0.020L \times 0.100M}}{{0.060L}} = 0.03333M\end{array}\)

	\({C_4}{H_4}{N_2}{O_3}(aq)\)	\(O{H^ - }(aq) \to \)	\({C_4}{H_4}{N_2}{O_3}^ - (aq) + \)	\({H_2}O(aq)\)
Initial (M)	0.06667	0.03333	0	/
Change (M)	-0.03333	-0.03333	+0.03333	/
Equilibrium (M)	-0.03333	0	+0.03333	/

Step:3

So, we are left with\(0.03333M{C_4}{H_4}\;{N_2}{O_3}and0.03333M{C_4}{H_3}\;{N_2}O_3^ - \)

	\(|{C_4}{H_{{4_{}}}}{N_2}{O_3}(aq)/rightleftheadpoons\)	\({H_3}{O^ + }\left( {aq} \right)\)	\({C_4}{H_4}{N_2}{O_3}^{ - )}(aq)\)
Initial (M)	0.03333	0	0.03333
Change (M)	-X	+X	+X
Equilibrium(M)	0.03333-X	X	0.03333+X

\(\begin{array}{l}{K_a} = \frac{{\left[ {1{1_3}{O^ + }} \right] \times \left[ {{C_4}{H_3}\;{N_2}{O_3}} \right]}}{{\left[ {{C_4}H{I_4}\;{N_2}{O_3}} \right]}}\\9.8 \times 1{0^{ - 5}} = \frac{{x \times (0.03333 + x)}}{{0.03333 - x}}\\ Since {K_a} is smaller than 1{0^4}, we will assume that 0.03333 + x\gg 0.03333 and 0.03333 - x\gg 0.03333\\9.8 \times 1{0^{ - 5}} = \frac{{x \times 0.03333}}{{0.03333}}\\x = 9.8 \times 1{0^{ - 5}}\\\left[ {{H_3}{O^ + }} \right] = 9.8 \times 1{0^{ - 5}}M\end{array}\)

Therefore, the \(pH\) value is

\(\begin{array}{c}pH = - log\left[ {{H_3}{O^ + }} \right]\\ = - log\left[ {9.8 \times 1{0^{ - 5}}} \right]\\ = 4.01\end{array}\)

To find the pH value in 39ml KOH solution added:

(c)

- We have\(0.040\;L\)of\(0.100M{C_4}{H_4}\;{N_2}{O_3}\)

- and\(39\;mL\)of\(0.100MKOH\)solution is added

- So, the new volume of a mixture is\(0.079\;L(79\;mL)\)

Let us calculate the\(pH\)value

The concentrations of\({C_4}{H_4}\;{N_2}{O_3}\)\(andKOH\)in a mixture is

\(\begin{array}{l}\left[ {{C_4}{H_4}{N_2}{O_3}} \right] = \frac{{0.040\;L \times 0.100M}}{{0.079\;L}} = 0.0506M\\(KOH] = \frac{{0.039\;L \times 0.100M}}{{0.079\;L}} = 0.04937M\end{array}\)

	\(|{C_4}{H_{{4_{}}}}{N_2}{O_3}(aq) \to \)	\({H_3}{O^ + }\left( {aq} \right)\)	\({C_4}{H_4}{N_2}{O_3}^{ - )}(aq)\)	\({H_2}O(aq)\)
Initial (M)	0.0506	0.04937	0	/
Change (M)	-0.04937	-0.04937	+0.04937	/
Equilibrium(M)	0.00127	0	0.4937	/

Step 5: 

So, we are left with\(0.00127M{C_4}{H_4}\;{N_2}{O_3}and0.04937M{C_4}{H_3}\;{N_2}O_3^ - \)

	\(|{C_4}{H_{{4_{}}}}{N_2}{O_3}(aq)/rightleftheadpoons\)	\({H_3}{O^ + }\left( {aq} \right)\)	\({C_4}{H_4}{N_2}{O_3}^{ - )}(aq)\)
Initial (M)	0.00127	0	0.004937
Change (M)	-X	+X	+X
Equilibrium(M)	0.00127-X	X	0.04937+X

\(\begin{array}{l}{K_a} = \frac{{\left[ {{H_3}{O^ + }} \right] \times \left[ {C{H_4}{H_3}\;{N_2}{O_3}} \right]}}{{\left[ {{C_4}{H_4}\;{N_2}{O_3}} \right]}}\\9.8 \times 1{0^{ - 5}} = \frac{{x \times (0.04937 + x)}}{{0.00127 - x}}Since{K_a}issmallerthan1{0^4}wewillassumethan,\\0.04937 + x\gg 0.04937and0.00127 - x\gg 0.00127\\9.8 \times 1{0^{ - 5}} = \frac{{x \times 0.04937}}{{0.00127}}\\x = 2.52 \times 1{0^{ - 6}}\\\left| {{H_3}{O^ + }} \right| = 2.52 \times 1{0^{ - 6}}M\end{array}\)

Therefore, the \(pH\) value is

\(\begin{array}{c}pH = - log\left[ {{H_3}{O^ + }} \right]\\ = - log\left[ {2.52 \times 1{0^{ - 6}}} \right]\\ = 5.60\end{array}\)

To find the PH value of 40ml of KOH solution added:

- We have\(0.040\;L\)of\(0.100M{C_4}{H_4}\;{N_2}{O_3}\)

- and\(40\;mL\)of\(0.100MKOH\)solution is added

- So, the new volume of a mixture is\(0.080\;L(80\;mL)\)

Let us calculate the\(pH\)value

The concentrations of\({C_4}{H_4}\;{N_2}{O_3}\)and\(KOH\)in a mixture is

\(\begin{array}{l}\left[ {{C_4}{H_4}\;{N_2}{O_3}} \right] = \frac{{0.040\;L \times 0.100M}}{{0.080L}} = 0.05\\(KOH] = \frac{{0.040\;L \times 0.100M}}{{0.080L}} = 0.05M\end{array}\)

	\(|{C_4}{H_{{4_{}}}}{N_2}{O_3}(aq) \to \)	\({H_3}{O^ + }\left( {aq} \right)\)	\({C_4}{H_4}{N_2}{O_3}^{ - )}(aq)\)	\({H_2}O(aq)\)
Initial (M)	0.05	0.05	0	/
Change (M)	-0.05	-0.05	0	/
Equilibrium(M)	0	0	0.05	/

Step 7: 

So, we are left with\(0.05M{C_4}{H_3}\;{N_2}O_3^ - \)

-\({K_b}of{C_4}{H_3}\;{N_2}O_3^ - is{K_b} = \frac{{{K_{11}}}}{{{K_a}}} = \frac{{1.1{0^{ - 14}}}}{{9.8 \times 1{0^{ - 3}}}} = 1.02 \times 1{0^{ - 10}}\)

	\(|{C_4}{H_{{4_{}}}}{N_2}{O_3}(aq)/rightleftheadpoons\)	\({H_3}{O^ + }\left( {aq} \right)\)	\({C_4}{H_4}{N_2}{O_3}^{ - )}(aq)\)
Initial (M)	0.05	0	0
Change (M)	-X	+X	+X
Equilibrium(M)	0.05	X	X

\(\begin{array}{l}{K_b} = \frac{{\left[ {{C_1}{H_1}{N_2}{O_3}} \right] \times [OH]}}{{\left[ {{C_4}{H_3}{N_2}O_3^ - } \right]}}\\1.02 \times 1{0^{ - 10}} = \frac{{x \times x}}{{0.05 - x}} Since {K_a} is smaller than 1{0^4} : we will assume that 0.05 + x\gg 0.05\\1.02 \times 1{0^{10}} = \frac{{x \times x}}{{0.05}}\\x = 2.26 \times 1{0^{ - 6}}\\\left[ {O{H^ - }} \right] = 2.26 \times 1{0^{ - 6}}M\end{array}\)Therefore, the\(pH\)value is

\(\begin{array}{c}pOH = - log\left[ {O{H^ - }} \right]\\ = - log\left[ {2.26 \times 1{0^{ - 6}}} \right]\\ = 5.65\\pH = 14 - 5.65\\ = 8.35\end{array}\)

To find the pH value of 41ml of KOH solution added:

- We have\(0.040\;L\)of\(0.100M{C_4}{H_4}\;{N_2}{O_3}\)

- and\(41\;mL\)of\(0.100MKOH\)solution is added

- So, the new volume of a mixture is\(0.081\;L(81\;mL)\)

Let us calculate the\(pH\)value

The concentrations of\({C_4}{H_4}\;{N_2}{O_3}\) and\(KOH\)in a mixture is

\(\begin{array}{l}\left[ {{C_4}{H_4}\;{N_2}{O_3}} \right] = \frac{{0.040\;L \times 0.100M}}{{0.081\;L}} = 4.938 \times 1{0^{ - 2}}M\\(KOH] = \frac{{0.041\;L \times 0.100M}}{{0.081L}} = 5.062 \times 1{0^{ - 2}}M\end{array}\)

	\(|{C_4}{H_{{4_{}}}}{N_2}{O_3}(aq) \to \)	\({H_3}{O^ + }\left( {aq} \right)\)	\({C_4}{H_4}{N_2}{O_3}^{ - )}(aq)\)	\({H_2}O(aq)\)
Initial (M)	\(4.938.1{0^{ - 2}}\)	0.05	0	/
Change (M)	-\(4.938.1{0^{ - 2}}\)	-\(4.938.1{0^{ - 2}}\)	+\(4.938.1{0^{ - 2}}\)	/
Equilibrium(M)	0	\(1.24.1{0^{ - 3}}\)	\(4.938.1{0^{ - 2}}\)	/

Now we have an excess of\(KOH\)

Therefore, the\(pH\)of a mixture is

\(\begin{array}{c}pOH = - log\left[ {O{H^ - }} \right]\\ = - log\left[ {1.24 \times 1{0^{ - 3}}} \right]\\ = 2.91\\pH = 14 - 2.91\\ = 11.09\end{array}\)