Question

Consider a buffer solution that contains 0.55 M NH₂CH₂CO₂H and 0.35 M NH₂CH₂CO₂Na. pK_a(NH₂CH₂CO₂H)=9.88.

a)Calculate the pH.

b)Calculate the change in pH if 0.155 g of solid NaOH is added to 250 mL of this solution.

c)If the acceptable buffer range of the solution is ±0.10 pH units, calculate how many moles of H₃O⁺ can be neutralized by 250 mL of the initial buffer.

Answer 1

a)

this is a buffer, so

pH = pKa + log(NH2CH2CO2Na / NH2CH2CO2H)

pH = 9.88 + log(0.35/0.55)

pH = 9.68

b)

mol of NaOH = mass/MW = 0.155/40 = 0.003875 mol of NaOH

mol of NH2CH2CO2H = MV = 0.55*0.250 = 0.1375

mol of NH2CH2CO2Na = MV =  0.35*0.250 = 0.0875

now

after neutralizaiton with NaOH

mol of NH2CH2CO2H = 0.1375-0.003875 = 0.133625

mol of NH2CH2CO2Na = 0.0875+0.003875 = 0.091375

pH = pKa + log(NH2CH2CO2Na / NH2CH2CO2H)

pH = 9.88+ log(0.091375/ 0.133625)

pH = 9.7149

c)

if only 0.1 pH unit:

mol of NH2CH2CO2H = 0.1375+x

mol of NH2CH2CO2Na = 0.0875-x

pH = pKa + log(NH2CH2CO2Na / NH2CH2CO2H)

pH = 9.88+ log((0.0875-x)/(0.1375+x))

9.88-0.10 = 9.88+ log((0.0875-x)/(0.1375+x))

10^(-0.10) = 0.79432 = (0.0875-x)/(0.1375+x)

(0.0875-x)/(0.1375+x)) = 0.79432

0.79432 *0.1375+ 0.79432 x = 0.0875-x

(1+0.79432 )x = 0.79432 *0.1375-0.0875

x = ( 0.79432 *0.1375-0.0875) / (1+0.79432 )

x = 0.012104

mol of H3O+ = 0.012104