More info-How do you calculate the expected change in pH of a buffer solution when acid/base is added?

I'm sorry, I forgot:

both NaOH and HCl are 0.5mol/L

Ka of ethanoic acid is 1.7 * 10^-5...

does that help?

here's the question again



A buffer of pH 4 is prepared by adding 0.14g of solid sodium ethaonate to 0.1l of 0.1mol/l ethanoic acid.

What will be the new pH if 2.2 mL of HCl are added?

If 4.5mL of NaOH are added?More info-How do you calculate the expected change in pH of a buffer solution when acid/base is added?
molar mass CH3COONa = 82

0.14 / 82 = 0.00171 moles CH3COONa

Moles CH3COOH = 0.01

We add 2.2 mL of HCl 0.5 M

Moles H+ = 0.5 x 2.2 /1000 = 0.0011

The effect of the added 0.0011 mole of H+ would be to decrease the moles of CH3COO- by 0.0011 and increase the moles of CH3COOH by 0.0011 by the reaction

H+ + CH3COO- %26gt;%26gt; CH3COOH

This will give us 0.00171 - 0.0011 = 0.00061 moles CH3COO- and 0.01+ 0.0011 = 0.0111 moles CH3COOH

Total volume = 0.1 L + 0.0022 L = 0.1022 L

[CH3COO-] = 0.00061 / 0.1022 = 0.00597 M

[CH3COOH ] = 0.0111 / 0.1022 = 0.1086 M

pK = - log 1.7 x 10^-5 = 4.77

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

= 4.77 + log 0.00597 / 0.1086 = 3.510



MOles OH- = 4.5 x 0.5 /1000 = 0.00225

The effect of the added OH- is to convert CH3COOH to CH3COO- via the net reaction

OH- + CH3COOH %26gt;%26gt; H2O + CH3COO-

The initial buffer has 0.00171 moles CH3COO- and 0.01 moles CH3COOH

We get 0.01 - 0.00225 = 0.0075 mole CH3COOH and

0.00171 + 0.00225 = 0.00396 mole CH3COO-

total volume = 0.0045 L + 0.1 = 0.1045 L

[ CH3COOH ] = 0.0075 / 0.1045 = 0.0718 M

[CH3COO-] = 0.00396 / 0.1045 = 0.0379 M

pH = 4.77 + log 0.0379 / 0.0718 = 4.49