consider the following compounds: NH4Cl, KHSO4, KNO2 and NaNO3... which compound

Question

consider the following compounds: NH4Cl, KHSO4, KNO2 and NaNO3... which compound would you choose to produce an aqueous solution of pH 8.65? and calculate the molarity of the solution that would yield a pH of 8.65? consider the following compounds: NH4Cl, KHSO4, KNO2 and NaNO3... which compound would you choose to produce an aqueous solution of pH 8.65? and calculate the molarity of the solution that would yield a pH of 8.65? consider the following compounds: NH4Cl, KHSO4, KNO2 and NaNO3... which compound would you choose to produce an aqueous solution of pH 8.65? and calculate the molarity of the solution that would yield a pH of 8.65?

Explanation / Answer

We would choose KNO2 for this solution to get a pH of 8.65.

Looking at the other compounds, NH4Cl is acidic, KHSO4 will be neutral salt, NaNO3 will again be a neutral salt. So the best option here is KNO2 basic in nature which will ionize in water to give a solution of pH 8.65 as desired.

Calculating the molarity of the solution that would yield a pH of 8.65

Ka = [H+][NO2-] / [HNO2]

[H+] = antilog(-8.65) = 2.24 x 10^-9 M (a basic solution)

[OH-] = Kw / [H+]

[OH-] = (1 x 10^-14)/(2.24 x 10^-9) = 4.47 x 10^-6 M

Ka = [H+][NO2-] / [HNO2] = 4.5 x 10^-4 (from standard literature)

Kb = Kw / Ka = (1 x 10^-14) / (4.5 x 10^-4) = 2.22 x 10^-11

The basic solution is formed by the hydrolysis of NO2-. The equilibrium constant is the same as the Kb:

NO2-(aq) + H2O <=> HNO2 + OH-(aq)

Kb = [HNO2][OH-] / [NO2-]

assuming [HNO2] = [OH-],

2.22 x 10^-11 = [4.47 x 10^-6][4.47 x 10^-6] / ([NO2-]-[4.47x10^-6])

or,

2.22x10^-11 = [4.47x10^-6]^2 / [NO2-]

since 4.47 x 10^-6 is very small compared to [NO2-].

Solve for [NO2-] = 0.90 M

Thus, KNO2 molarity would be 0.90 M

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