Question

help with 2 buffer questions please.
data table and questions in pics thank you

Buffer B Buffer A .0 39 3 Mass of NaC2H3O2 used to prepare buffer (g) NO0-0 20 Volume of buffer prepared (mL) 1.0 0.1 Molar concentration of HC2H3O2 in buffer (M) 4.13 4.23 3.7 Initial pH of buffer Volume of 0.5 M NAOH to raise pH by 2 units (mL) 37.5 4.2mL Volume of 0.5 M HCI to lower pH by 2 units (mL) 3,1 m35 nl Volume of 0.5 M NaOH at equivalence point (mL)

Answer 1

1. Buffer C

Added 8.203 g of Sodium acetate (CH₃COONa )

Moles of CH₃COONa =    8.203 g / 82.03 g/mol = 0.1 mol

  CH₃COOH : 100 ml of 1.0 M

Concentration of CH₃COONa =  0.1 mol / 100 ml = 0.1 mol / 0.1 L = 1.0 M

pH of buffer solution is calculated using this equation :

  pH = pKa + log ((A^-) /(HA)) (HA - weak acid ; A^- - conjugated base of HA)

Initial pH of this buffer ;

pH = pKa + log  ((CH₃COONa) /(CH₃COOH))

pka for Acetic acid = 4.74

pH = 4.74 + log  ((1.0 M) /1.0 M )) = 4.74

2.   pH change on Adding 5.0 ml of 0.5 NaOH in 20.0 ml of buffer B and C.

addition of NaOH leads to reaction :

CH₃COOH(aq) + OH^- (aq) -----> CH₃COO^- (aq) + H2O(l)

It increase amount of CH₃COONa .

We Calculate change in concentrations and then pH :

For Buffer C :

(CH₃COOH) = moles unreacted CH₃COOH /total volume

= (20 ml*1.0 M) - (5 ml *0.5 M ) / 25 ml = 0.7 M

(CH₃COONa) = (20 ml*1.0 M) + (5 ml *0.5 M )/ 25 ml = 0.9 M

pH = 4.74 + log  ((0.9) /(0.7)) = 4.85

For Buffer B   :

initial pH of buffer : 4.13

initial (CH₃COOH) = 1.0 M

initial (CH₃COONa) = (0.39g / 82 g/mol ) / 0.020 L = 0.24 M

We Calculate change in concentrations and then pH :

(CH₃COOH) = moles unreacted CH₃COOH /total volume

= (20 ml*1.0 M) - (5 ml *0.5 M ) / 25 ml = 0.7 M

(CH₃COONa) = (20 ml*0.24 M) + (5 ml *0.5 M )/ 25 ml = 0.29 M

pH = 4.74 + log  ((0.29) /(0.7)) = 4.36

So, change in pH for buffer C is 0.11 and for buffer B is 0.23. ; for buffer C change in pH is less.