Question

What mass of NaCH₃CO₂ (MW: 82.034g/mol) must be dissolved in 0.300 L of 0.500 M HCH₃CO₂ (Ka= 1.8x10^-5) to produce a solution with pH=5.55? Assume the volume remains constant.

Answer 1

Using Henderson-Hasselbalch equation, we get

pH = 5.55,

Ka= 1.85 x 10^-5

pKa = - log(1.85 x 10^-5) = 4.73

[CH₃COOH] = 0.5 M

Let the concentration of [CH₃COONa] = m

So, substituting the values in henderson-Hasselbalch equation, we get

5.55 = 4.73 + log (m/0.5)

log (m/0.5) = 0.82

m/0.5 = 10^0.82

m/0.5 = 6.6

m = 6.6 x 0.5 = 3.30

So, m = [CH₃COONa] = 3.3 M

3.3 M of CH₃COONa means

In 1 L = 3.3 moles of CH₃COONa

0.3 L = (0.3 x 3.3) moles of CH₃COONa

0.3 L = 0.99 moles of CH₃COONa

Now, 1 mole of  CH₃COONa = 82.034 g

0.99 moles of CH₃COONa = (0.99 x 82.034) g = 81.214 g