how many grams of sodium acetate, NaC_2H_3O_2, must be used to make 250.0 mL of

Question

how many grams of sodium acetate, NaC_2H_3O_2, must be used to make 250.0 mL of an aqueous solution having a pH of 8.50? For HC_2H_3O_2, K_a = 1.8 times 10^-5. What is the pH of a 0.150 M solution of potassium cyanide, KCN? Hydrocyanic acid, HCN, is a weak acid for which K_a = 4.93 times 10^-10. Sodium benzoate is the salt of benzoic acid, a weak acid with K_a = 6.6 times 10^-5. What is the pH of a 0.25 M solution of sodium benzoate? What is the molarity of a potassium acetate solution that has a pH of 9.05? (See question 4 for additional information.) potassium sorbate is used as a preservative in some foods, including cheese. it is a salt of sorbic acid,HC_6H_7O_2,a week acid for which K_a = 1.7 times 10^-5.what is the pH of a solution formed by dissolving 11.25 g of potassium sorbate in 1.75 L of solution?

Explanation / Answer

4. pH = 8.50

pOH = 14 - pH = 5.5

[OH-] = 3.16 x 10^-6 M

For sodium acetate,

Kb = 1 x 10^-14/1.8 x 10^-5 = [HC2H3O2][OH-]/[C2H3O2-]

[OH-] = [HC2H3O2]

we get,

[C2H3O2-] = (3.16 x 10^-6)^2/5.55 x 10^-10 = 0.018 mol/L

So, grams of sodium acetate to be added = 0.018 mol/L x 0.250 L x 82.03 g/mol = 0.37 g

7. Molarity of KC2H3O2

pH = 9.05

pOH = 14 - pH = 4.95

[OH-] = 1.12 x 10^-5 M

For potassium acetate,

Kb = 1 x 10^-14/1.8 x 10^-5 = [HC2H3O2][OH-]/[C2H3O2-]

[OH-] = [HC2H3O2]

So the molarity of solution would be,

[C2H3O2-] = (1.12 x 10^-5)^2/5.55 x 10^-10 = 0.226 M

8. molarity of KC6H7O2 solution = 11.25 g/150.22 g/mol x 1.75 L = 0.043 M

sorbate hydrolyses as,

C6H7O2- + H2O <==> HC6H7O2 + OH-

let x amount has reacted

Kb = 1 x 10^-14/1.7 x 10^-5 = x^2/0.043

x = [OH-] = 5.03 x 10^-6 M

pOH = -log[OH-] = 5.30

So pH of solution is,

pH = 14 - pOH = 8.70

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