Calculate the pH of a solution that is 0.8M acetic acid and 0.9 M sodium acetate

Question

Calculate the pH of a solution that is 0.8M acetic acid and 0.9 M sodium acetate after the addition H_2 SO_4 The pKa of acetic acid is 4.75. Fe^3+ (aq) + SCN^- (aq) FeSCN^2+ (aq) What is the equilibrium constant expression Kc for the above reaction? A student does an experiment to determine the equilibrium constant for the same reaction that from Problem 12, Fe^3+ (aq) + SCN^- (aq) FeSCN^2+ (aq) The student mixes 5.00 mL of 2.00 times 10^-3 M Fe(NO_3)_3 solution with 5.00 mL of 2.00 times 10^-3 M KSCN solution, heats the mixture, and finds that the equilibrium concentration of FeSCN^2+ in the mixture is 5.00 times 10^-5 M. Calculate the equilibrium constant for the reaction under the conditions in this experiment

Explanation / Answer

Q12.

Kc = [procuts]^p /[reactants]^r

so

Kc = [FeSCN+2]/ [Fe+3][SCN-]

Q13.

Vtotal = 5+5 =10

[Fe+3] = mol*V /Vtotal = 5/10*2*10^-3 = 10^-3 M of Fe+3

[SCN-] = mol*V /Vtotal = 5/10*2*10^-3 = 10^-3 M of SCN-

[FeSCN-] = 0

then... in equilibrium

[Fe+3] = 10^-3 M of Fe+3 - x

[SCN-] = 10^-3 M of SCN- - x

[FeSCN-] = 0 + x

so..

[FeSCN-] = 0 + x = 5*10^-5

x =  5*10^-5

[Fe+3] = 10^-3 -  5*10^-5 = 0.00095

[SCN-] = 10^-3 - 5*10^-5 = 0.00095

K = 0.00095*0.00095/( 5*10^-5)

K = 0.01805

K = 1.8*10^-2

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