9. If enough of a monoprotic acid is dissolved in water to produce a 0.018 M sou

Question

9. If enough of a monoprotic acid is dissolved in water to produce a 0.018 M soultion with a pH of 6.58, what is the equilibrium constant, Ka, for the acid?

10. For the following chemical reaction 2HBr(aq)+ Ba(OH)2(aq)----> 2H2O(l) + BaBr (aq) write the net ionic equation, including the phrases.

11. For the following chemical reaction HCN(aq) + KOH(aq) ---> H2O(l)+ KCN(aq) write net ionic equation, including the phrases.

12. The flask shown here contains 10.0 Ml of HCl and a few drops of phenolphthalein indicator. The buret contains 0.100 M NaOh.

- What volume of NaOH is needed to reach the end point of the tritration?

- What was the initial concentration of HCl?

14.a) What is the pH of an aqueous solution with a hydrogen ion concentration of [H+]= 2.1 * 10^-9 M?

     b) What is the hydroxide ion concentration, [OH-], in an aqueous solution with a hydrogen ion concentration of [H=]= 2.1* 10^-9 M?

     c) A monoprotic acid, HA, dissociates:

        HA---> H^+ + A^-

The equilibrium concentrations of the reactant and products are:

[HA]=0.180M

[H+]=2.00810^-4M

[A^-]=2.00810^-4M

calculate the Ka value for the acid HA.

Explanation / Answer

9)

let the acid be HA

HA <-------> H+ (aq) + A- (aq)
0.018 0 0 (initial)
0.018-x x x (at equilibrium)

pH = -log [H+]
6.58 = -log [H+]
[H+] = 2.63*10^-7 M
[H+] = x = 2.63*10^-7
Ka = [H+][A-] /[HA]
= x*x / (0.0146-x)
= (2.63*10^-7)*(2.63*10^-7) / (0.018 - 2.63*10^-7)
2.63*10^-7 is small and can be ignored as compared to 0.018
above expression thus becomes,
Ka = (2.63*10^-7)*(2.63*10^-7) / (0.018 )
= 3.84*10^-12

Answer: 3.84*10^-12

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