Question

Calculate the pH of a 1 L solution that contains 1.0 M HCl and 1.5 M CH3COOK. The Ka for CH3COOH is 1.8 x 10-5

Answer 1

we know that

moles = molarity x volume (L)

so

moles of HCl = 1 x 1= 1

moles of CH3COOK = 1.5 x 1= 1.5

now

the reaction is

H+ + CH3COO- = CH3COOH

now

moles of CH3COO- reacted = moles of H+ added = 1

moles of CH3COOH formed = moles of H+ added = 1

so

finally

moles of CH3COO- = 1.5 - 1 = 0.5

moles of CH3COOH = 1

now

molarity = moles / volume (L)

so

[CH3COO-] = 0.5 / 1 = 0.5

[CH3COOH] = 1 / 1= 1

now

CH3COOH and CH3COO- form a buffer solution

for buffers

pH = pKa + log [salt / acid ]

aslo

pKa = -log Ka

so

pH = -logKa + log [CH3COO- / CH3COOH]

so

pH = -log 1.8 x10-5 + log [ 0.5 / 1]

pH = 4.443

so

the pH of the solution is 4.443