Question

For 490.0 mL of a buffer solution that is 0.125 M in HC2H3O2 and 0.110 Min NaC2H3O2, calculate the initial pH and the final pH after adding 0.010 mol of HCl.

Express your answers using two decimal places separated by a comma.

Answer 1

we know that

for buffers

pH = pKa + log [salt / acid ]

so

pH = pKa + log [CH3COONa / CH3COOH}

so

pH = 4.76 + log [ 0.11 / 0.125]

pH = 4.7045

so

the pH before addition oos HCl is 4.7045

now

we know that

moles = molarity x volume (L)

so

moles of CH3COOH = 0.125 x 0.490 = 0.06125

moles of CH3COONa = 0.11 x 0.490 = 0.0539

now

moles of HCl added = 0.01

the reaction is

CH3COO- + H+ ---> CH3COOH

so

moles of CH3COO- reacted = moles of HCl added = 0.01

moles of CH3COOH formed = moles of HCl added = 0.01

now

finally

moles of CH3COO- = 0.0539 - 0.01 = 0.0439

moles of CH3COOH = 0.06125 + 0.01 = 0.07125

now

pH = pKa + log [ CH3COONa / CH3COOH]

pH = 4.76 + log [ 0.0439 / 0.07125]

pH = 4.55

so

the final pH of the solution is 4.55