Question

A volume of 500.0 mL of 0.150 M NaOH is added to 615 mL of 0.250 M weak acid (Kg = 4.65 x 10-'). What is the pH of the resulting buffer? HA(aq) + OH(aq) — H,O(l) + A (aq) pH

Answer 1

1)

Given:

M(HA) = 0.25 M

V(HA) = 615 mL

M(NaOH) = 0.15 M

V(NaOH) = 500 mL

mol(HA) = M(HA) * V(HA)

mol(HA) = 0.25 M * 615 mL = 153.75 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.15 M * 500 mL = 75 mmol

We have:

mol(HA) = 153.75 mmol

mol(NaOH) = 75 mmol

75 mmol of both will react

excess HA remaining = 78.75 mmol

Volume of Solution = 615 + 500 = 1115 mL

[HA] = 78.75 mmol/1115 mL = 0.0706M

[A-] = 75/1115 = 0.0673M

They form acidic buffer

acid is HA

conjugate base is A-

Ka = 4.65*10^-5

pKa = - log (Ka)

= - log(4.65*10^-5)

= 4.333

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.333+ log {6.726*10^-2/7.063*10^-2}

= 4.311

Answer: 4.31

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