A 25.00 mL sample of 0.260 M NaOH analyte was titrated with 0.750 M HCl at 25 °C

Question

A 25.00 mL sample of 0.260 M NaOH analyte was titrated with 0.750 M HCl at 25 °C.

Calculate the initial pH before any titrant was added.

pH=

Calculate the pH of the solution after 5.00 mL of the titrant was added.

pH=

2. The half-equivalence point of a titration occurs half way to the end point, where half of the analyte has reacted to form its conjugate, and the other half still remains unreacted. If 0.560 moles of a monoprotic weak acid (ka=8.6x10^-5) is titrated with NaOH, what is the pH of the solution at the half equivilance point?

pH=

3.Calculate the pH at the equivalence point for the titration of 0.240 M methylamine (CH3NH2) with 0.240 M HCl. The Kb of methylamine is 5.0× 10–4.

Explanation / Answer

a)
before adding any titrant
[NaOH] = [OH-] = 0.260 M
pOH = -log [OH-]
= -log (0.260)
= 0.585
pH = 14 - pOH
= 14 - 0.585
= 13.415
Answer: 13.4

b)
When 5 mL of HCl is added,
rmaining moles of NaOH = 0.260*0.025 - 0.750*0.005
=2.75*10^-3 mol

Total volume = 0.025 + 0.005 = 0.03 L

[NaOH] = [OH-] = (2.75*10^-3)/0.03 = 0.092 M

pOH = -log [OH-]
= - log (0.092)
= 1.04

pH = 14 -pOH
= 14 - 1.04
=12.96
Answer: 12.96

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