Question

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

1) Calculate its pH

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

3) 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

Q1

this is a buffer so:

pH = pKa + log(base/acid)

pH = 9.88 + log(0.35/0.55)

pH = 9.6837

b)

mol of acid = MV = 0.35*0.25 = 0.0875

mol of conjguate = MV = = 0.55*0.25 = 0.1375

after adding base:

mol of base = MV = 0.155/40 = 0.003875

mol of acid = = 0.0875-0.003875 = 0.083625

mol of conjguate == 0.1375+0.003875 = 0.141375

pH = 9.88 + log(0.141375/0.083625)

pH = v

c)

if pH drop is 0.1

then

initial pH = 9.6837

pH drop max = pH = 9.6837-0.1 = 9.5837

pH = pKa + log(base/acid)

9.5837 = 9.88 + log((0.1375-x)/(0.0875+x))

10^(9.5837-9.88 ) = (0.1375-x)/(0.0875+x)

0.5054*(0.0875+x) = 0.1375-x

0.04422 + 0.5054x = 0.1375-x

(1+0.5054)x = 0.1375-0.04422

x = (0.1375-0.04422 )/(1+0.5054)

x = 0.06196 mol of acid