Question

Calculate the [H₃O⁺] and pH of each polyprotic acid solution. (See the appendix.)

(a) 0.272 M H₂CO₃

[H3O+]	M
pH

(b) 0.133 M H₃C₆H₅O₇

[H3O+]	M
pH

Answer 1

Ans a) H₂CO₃ is a weak acid

H₂CO₃ +H₂O $\rightleftharpoons$ H₃O⁺ +HCO₃^-

	H2CO3	H3O+	HCO3-
initial	0.272	0	0
change	- x	x	x
equilibrium	0.272-x	x	x

Here K_a = 4.3 x 10^-7

$K_{a}=\frac{[H_{3}O^{+}][HCO_{3}^{-}]}{[H_{2}CO_{3}]}$

4.3 x 10^-7 = x² / 0.272-x

x=0.645

[H₃O⁺] = 0.645

pH= - log[H₃O⁺]

    = 3.89

Ans;- b) H₃C₆H₅O₇ + H₂O $\rightleftharpoons$ H₃O⁺ + C₆H₅O₇^-

Here K_a = 6.46 x 10^-5

	H3C6H5O7	H3O+	C6H5O7-
initial	0.133	0	0
change	- x	+x	+x
equilibrium	0.133-x	x	x

$K_{a}= \frac{[H_{3}O^{+}][C_{6}H_{5}O_{7}^{-}]}{H_{3}C_{6}H_{5}O_{7}}$

6.46 x 10^-5 = x² / 0.133- x

x = 0.00110

[H₃O⁺] = 0.00110

pH= - log[H₃O⁺]

    = 2.95