hey can you answer this 3 please and thank you!!!!

Question

17).Calculate the pH of a 0.250 M Ba(OH)₂ solution at 25^oC?

19.)   Calculate the pH of a 0.250 M solution of benzoic acid (C₆H₅CO₂H) if  

 K_a = 6.3 x 10^-5 for the acid.   C₆H₅CO₂H(aq)    ⇌   H⁺(aq) + C₆C₅CO₂^-(aq)

20.)  a) Calculate the pH of a 0.150 M solution of formic acid, HCO₂H? 

    K_a, HCO₂H = 1.8 x 10^-4.   HCO₂H(aq) ⇌ HCO₂^-(aq) + H⁺(aq)

                       ____________________________

 b) What is the concentration of HCO₂^- (aq) at equilibrium?   _____________

Answer 1

17) 0.250 M Ba(OH)2 ==> 0.250 M Ba2+ + 0.50 M OH-

pOH = - log [OH-] = -log 0.50 = 0.30

pH = 14 - pOH

pH = 13.7

19) Ka = 6.3x10^-5 = [H⁺][C₆H₅CO₂^-]/[C₆H₅CO₂H]

6.3x10^-5 = (x)(x)/0.250 - x (and assuming x is small, it can be neglected in the denominator)

x² = 1.58x10^-4

x = 1.26x10^-2 M = [H⁺] (this is about 5% of 0.250 so right on the borderline of being neglected)

pH = -log [H⁺] = -log 1.26x10^-2

pH = 1.90

20) Ka = 1.8x10^-4 = [H⁺][HCO₂^-]/[HCO₂H]

a) 1.8x10^-4 = (x)(x)/0.150 - x (assume x is small and neglect it in the denominator)

x² = 2.7x10^-5

x = 5.2x10^-3 = [H⁺] (this is less than 4% of 0.150 M so assumption was valid)

pH = -log 5.2x10^-3

pH = 2.28

[HCO₂^-]@ equilibrium = 5.2x10^-3 M (same as equilibrium concentration of H⁺)