Question

What is the pH of 75 mL 1.2 M NaOH mixed with 50 mL 0.90 M HCl?

A buffer is prepared by mixing 200.0 mL of 0.1500 M NaOH with 200.0 mL of 0.200 M CH₃CO₂H in a 1-L volumetric flask and then diluted to volume with distilled deionized water. Calculate the pH of the buffer. K_a = 1.75 × 10⁻⁵

Answer 1

Q1. What is the pH of 75 mL 1.2 M NaOH mixed with 50 mL 0.90 M HCl?

strong acid + strong base = water + salt

mmol of NaOH = MV = 1.2*75 = 90 mmol

mmol of HCl = MV = 50*0.9 = 45 mmol

therefore,

NaOH > HCl, then

mmol of NAOH left = 90-45 = 45 mmol

[NaOH] = mmol of NaOH / Vtotal = (45)/(75+50) = 0.36

pOH = -log([OH-]) = -log(0.36) = 0.44369

pH = 14-pOH = 14-0.44369

pH = 13.55631

Q".

the buffer equation is given by henderson haselbachs equation

pH = pKa + log(A-/HA)

in this case, A- = CH3COO- and CH3COOH is the HA acid

pKA = -log(KA) = -log(1-8*10^-5) = 4.75

now,

mmol of weak acid = MV = 0.2*1000 = 200

mmol of strong base = MV = 200*0.15 = 30

after reaction, there is neutralization, weak acid reacts, and conjugate base forms:

mmol of HA = 200-30 = 170 left

mmol of A- formed = 0 + 30 = 30

now, substitute

pH = pKa + log(A-/HA)

pH = 4.75+ log(30/170)

pH = 3.99667