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QUESTION

# Determine the pH of a .22-M NaF solution at 25^@"C"? The K_a of HF is 3.5xx10^(-5)

The of your solution will be equal to 8.90.

Sodium fluoride, NaF, is a soluble salt that dissociates completely in aqueous solution to give sodium cations, Na^(+), and fluoride anions, F^(-).

NaF_((s)) -> Na_((aq))^(+) + F_((aq))^(-)

Since 1 mole of sodium fluoride produces 1 mole of fluoride anions, the concentration of the fluoride ions will be equal to that of the salt.

[F^(-)] = [NaF] = "0.22 M"

The fluoride ions will react with water to form hydrofluoric acid, a weak acid, and hydroxide ions, OH^(-), which is an indicator that the of the solution will be greater than 7.

Use an ICE table to determine the of the hydroxide ions

" "F_((aq))^(-) + H_2O_((l)) rightleftharpoons HF_((aq)) + OH_((aq))^(-)I....0.22...............................0..................0C....(-x)...............................(+x)..............(+x)E...0.22-x............................x...................x

The base dissociation constant, K_b, will be equal to

K_b = K_w/K_a = 10^(-14)/(3.5 * 10^(-5)) = 2.86 * 10^(-10)

This means that you'll get

K_b = ([HF] * [OH^(-)])/([F^(-)]) = (x * x)/(0.22 - x) = x^2/(0.22-x)

Since K_b is so small, you can approximate "(0.22-x)" with 0.22 to get

K_b = x^2/0.22 = 2.86 * 10^(-10)

Therefore,

x = sqrt(0.22 * 2.86 * 10^(-10)) = 7.9 * 10^(-6)

You can now determine the pOH of the solution

pOH = -log([OH^(-)])

pOH = -log(7.9 * 10^(-6)) = 5.10

As a result, the solution's pH will be

pH_"sol" = 14 - pOH = 14 - 5.10 = color(green)("8.90")