Polyprotic Acids and Bases Questions. Please explain! thank you Problem 1: What

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Polyprotic Acids and Bases Questions. Please explain! thank you

Problem 1: What is the pH of a 0.10 M salution of oxalic acid, H.c,0. What are the cancentrations of H3O HC204, and the oxalate ion Czo.2 H20 (aq) Ka 6.4 x 10 Problem 2: What is the pH of a 0.10 M solution af sodium carbonate, Nazco at 25 C HCO's (aq) H1o (aq 4.2 x 10 CO2 -(aq) Hao (aq) Kala 4.8 x 10 Special Case: Dilute H2so4 solution one must take into account the contribution of K.2 to the solution pH. Find the pH of a o.0100 M solution of Hrso. solution.

Explanation / Answer

Q1.

pH of a 0.1 M of oxalic acid:

initially

H2C2O4 <-< H+ + HC2O4- Ka1

Ka1 = [H+][HC2O4-]/[H2C2O4]

[H+] = x= [HC2O4-] (due to stoichiometry)

[H2C2O4] = M-x = 0.1-x (accounting for dissociation )

5.9*10^-2 = x*x/(0.1-x)

solve for x

x = 0.05278

meaning that:

[H+] = x= 0.05278

[HC2O4-] = x = 0.05278

[H2C2O4] = 0.1-x= 0.04722

now,,

[HC2O4-] will dissociate as follows:

HC2O4- <-> H+ + C2O4-2 Ka2

Ka2 = [H+][C2O4-2]/[HC2O4]

intially:

[H+] = 0.05278

[C2O4-2] = 0

[HC2O4] = 0.05278

after equilibrium

[H+] = 0.05278 + y

[C2O4-2] = 0 + y

[HC2O4] = 0.05278 - y

so

6.4*10^-5 = (0.05278 + y)(y) / (0.05278 - y)

solve for y

(6.4*10^-5)(0.05278 ) - y*(6.4*10^-5) = 0.05278y + y^2

y^2 + y(0.05278 + 6.4*10^-5) - ((6.4*10^-5)(0.05278 ) ) = 0

y^2 + 0.052844y - 3.37792*10^-6 = 0

y = 6.36*10^-5

[H+] = x+y = 0.05278 +6.36*10^-5 = 0.0528436 M

so

pH = -log(H ) = -log(0.0528436) = 1.2770

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