Question

Consider the titration of 20.0 mL of 0.100M HCl with 0.200 M NaOH solution. Calculate the...

Consider the titration of 20.0 mL of 0.100M HCl with 0.200 M NaOH solution. Calculate the pH after the addition of the following volumes of sodium hydroxide solution.

A) 0.00 mL

B) 12.00 mL

C) 20.00 mL

D) 25.00 mL

Homework Answers

Answer #1

this is strong acid + strong base

then

initial mmol of acid = MV = 20*0.1 = 2

a)

[H+] = [HCl] = 0.1

pH = -log(0.1) = 1

b)

mmol of base = MV = 0.2*12 = 2.4

HCl + NaOH = H2O + NaCl

mmol of base left = 2.4 - 2 = 0.4

[NaOH] = mmol/V = 0.4 / (20+12) = 0.0125

pH = 14- pOH = 14 + log(OH) = 14+log(0.0125) = 12.09

c)

mmol of base = MV = 0.2*20= 4

HCl + NaOH = H2O + NaCl

mmol of base left = 4- 2 = 2

[NaOH] = mmol/V = 2/ (20+20) = 0.05

pH = 14- pOH = 14 + log(OH) = 14+log(0.05) = 12.69

d)

mmol of base = MV = 0.2*25= 5

HCl + NaOH = H2O + NaCl

mmol of base left = 5- 2 = 3

[NaOH] = mmol/V = 3/ (20+25) = 0.0666

pH = 14- pOH = 14 + log(OH) = 14+log(0.0666) = 12.82

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