Answer to Question #126460 in Chemistry for Bbbb

Question #126460
A 50.0mL solution containing 0.1M H2SO4 and 0.2M maleic acid is titrated with 0.5M NaOH , calculated the pH after each of the addition, a)zero b)10 mL c)20 mL d)30 mL e)40 mL f)50 g)55 h) 60 i)70 mL of NaOH. *
1
Expert's answer
2020-07-16T07:00:12-0400

Both sulfuric (H2SO4) and maleic acids are diprotic acids. Here are the deprotonation stages and corresponding pKa values:

H2SO4 "\\leftrightarrow" H+ + HSO4-, pKa1 = -3.0

HSO4- "\\leftrightarrow" H+ + SO4 2-, pKa2 = 1.9

H2M "\\leftrightarrow" H+ + HM-, pKa1 = 1.83

HM- "\\leftrightarrow" H+ + M2-, pKa2 = 6.07

We must also include the equation of the autoprotolysis of water and its equilibrium constant:

H2O "\\leftrightarrow" H+ + OH-, pKw = 14.

Therefore, we end up with 8 equations and 8 unknowns:

"10^3 = \\frac{[H^+][HSO_4^-]}{[H_2SO_4]}"

"10^{-1.9} = \\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}"

"10^{-1.83} = \\frac{[H^+][HM^{-}]}{[H_2M]}"

"10^{-6.07} = \\frac{[H^+][M^{2-}]}{[HM^-]}"

"\\frac{V_a\u00b7c(H_2SO_4)}{V_a+V_b} = [H_2SO_4]+[HSO_4^-]+[SO_4^{2-}]"

"\\frac{V_a\u00b7c(H_2M)}{V_a+V_b} = [H_2M]+[HM^-]+[M^{2-}]"

"[H^+] +\\frac{c_bV_b}{V_a+V_b} = [HSO_4^-]+2\u00b7[SO_4^{2-}] +[HM^-]+2\u00b7[M^{2-}] + [OH^-]"

"10^{-14} = [H^+][OH^-]" .

The first four equations and the last equation are the equations for the equilibrium constants for the dissociation of acids and the autoprotolysis of water. The fifth and sixth equations are the equations of the mass balance of the acids and the base added. The sixth equation is the charge balance. The volumes of the acid and the base are "V_a" and "V_b" , respectively. The concentration of the base is denoted as "c_b" . Solving these equations for different volumes of the base, we get the following values of hydrogen cation concentration and pH:

Vb c(H+), M pH

a) 0 mL 1.29E-01 0.89

b) 10 mL  5.26E-02 1.28

c)20 mL  1.92E-02 1.72

d)30 mL  6.04E-03 2.22

e)40 mL  8.54E-05 4.07

f)50 mL 8.51E-07 6.07

g)55 mL 2.84E-07 6.55

h) 60 mL 3.07E-10 9.51

i)70 mL 2.40E-13 12.62.

From this data, we can draw the titration curve:


N.B. The equations have been solved using the Excel solver utility.


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