Question

You combine 0.75 moles formate and 0.85 moles formic acid to make a buffer solution. The Ka of formic acid is 1.8xl0'4 what is the pH of the solution? With the same initial solution of 0.75 moles formate and 0.85 moles formic acid to make a buffer solution. The Ka of formic acid is 1.8xl0'4. You add 8g sodium hydroxide to this solution, what is the new pH? With the same initial solution of 0.75 moles formate and 0.85 moles formic acid to make a buffer solution. The Ka of formic acid is 1.8x10-4. You add 16g hydrobromic acid to this solution, what is the new pH? With the same initial solution of 0.75 moles formate and 0.85 moles formic acid to make a buffer solution. The Ka of formic acid is 1.8x10-4. You add 60.7g hydrobromic acid to this solution, what is the new pH?

Answer 1

Ans 1 assume 1 L of solution

[Acid] = 0.85 M;

[salt] = 0.75 M

pK_a = – log{(K_a) / M} = – log (1.8 x 10^-4) = 3.75

pH = pK_a + log[salt] / [acid]) = 3.75 + log(0.75M / 0.85M)

pH = 3.75 + log(0.88) = 3.75 - 0.056 = 3.69

Ans 2 -

assume 1 L of solution

Add 8 g NaOH

Moles of NaOH added = 8/40 = 0.2 moles

The concentration will change

[Acid] = 0.85 - 0.2 = 0.65 M;

[salt]  = 0.75 + 0.2 = 0.95 M

pK_a = – log{(K_a) / M} = – log (1.8 x 10^-4) = 3.75

pH = pK_a + log[salt] / [acid]) = 3.75 + log(0.95M / 0.65M)

pH = 3.75 + log(1.46) = 3.75 + 0.16 = 3.91

Ans 3

assume 1 L of solution

Add 16 g HBr

Moles of HBr added = 16/80.91 = 0.2 moles

The concentration will change

[Acid] = 0.85 + 0.2 = 1.05 M;

[salt]  = 0.75 - 0.2 = 0.55 M

pK_a = – log{(K_a) / M} = – log (1.8 x 10^-4) = 3.75

pH = pK_a + log[salt] / [acid]) = 3.75 + log(0.55M / 1.05M)

pH = 3.75 + log(0.52) = 3.75 - 0.28 = 3.47

Ans 4

assume 1 L of solution

Add 60.7 g HBr

Moles of HBr added = 60.7/80.91 = 0.75 moles

The concentration will change

[Acid] = 0.85 + 0.75 = 1.60 M;

[salt]  = 0.75 - 0.75 = 0 M

pK_a = – log{(K_a) / M} = – log (1.8 x 10^-4) = 3.75

pH = pK_a + log[salt] / [acid]) = 3.75 + log(0M / 1.60M)

pH = 3.75 + log(0) = 3.75 = 3.75