Question

A buffer contains 1.0 mol of CH3CO2H and 1.0 mol of CH3CO2- diluted with water to 1,0 L. How many moles of NaOH are required to increase the pH of the buffer to 5.10? pKa of CH2CO2H is 4.74

Can someone explain this step by step please!

Answer 1

Answer – We are given, moles of CH₃COOH = 1.0 moles

Moles of CH₃COO^- = 1.0 moles , volume of water = 1.00 L , pH = 5.10

pKa = 4.74

We know Henderson Hasselbalch equation –

pH = pKa + log [CH₃COO-] / [CH₃COOH]

5.10 = 4.74 + log [CH₃COO^-] / [CH₃COOH]

So, log [CH₃COO-] / [CH₃COOH] = 5.10 -4.74

                                                         = 0.36

Taking antilog from both side

[CH₃COO^-] / [CH₃COOH] = 2.29

We are given the initial [CH₃COO-] = 1.0 mole / 1.0 L = 1.0 M

[CH₃COOH] = 1.0 mole / 1.0 L = 1.0 M

We know the ratio of [CH₃COO^-] / [CH₃COOH] = 2.29 after the addition of NaOH

When we added the NaOH then there is increase the moles of conjugate base and decrease the moles of acid

So, [CH₃COO-] = 1.0 +x

[CH₃COOH] = 1.0-x

1.0+x / 1.0-x = 2.29

1.0+x = (1.0-x) 2.29

1.0+x = 2.29 -2.29x

2.29x+x = 2.29 – 1.0

3.29 x = 1.29

x = 1.29/3.29

    = 0.392 moles

Moles of NaOH = 0.392
so, 0.392 moles of NaOH are required to increase the pH of the buffer to 5.10