Your turn. Calculate the pH of the solution when the following substances are ad
ID: 500387 • Letter: Y
Question
Your turn. Calculate the pH of the solution when the following substances are added together: 20 mL of 0.001 M HCI and 40 mL of 1.5 M Acetic acid 20 mL of 0.001 M HCI and 50 mL of 2.5 M Sodium Acetate 20 mL of 0.001 M HCl and 0.5 mL of 0.04 M Sodium Acetate 200 ml of 0.1 M HCl and 8 mL of 2.5 M Sodium Acetate 20 mL of 0.1 M NaOH and 8 mL of 0.25 M Sodium Acetate 20 mL of o.1 M NaoH and 8 ml of o.25 M Acetic acid 200 mL of 0.1 M NaoH and 80 mL of 2.5 M Acetic Acid 200 mL of 0.1 M NaOH and 8 mL of 0.25 M Acetic Acid 200 mL of 0.01 M HCl and 8 ml of 0.25 NaoH 200 mL of 0.01 M Acetic Acid and 80 mL of 2.5 MSodium AcetateExplanation / Answer
Number of moles = molarity (mol/L) * Volume (L)
1. nHCl = 0.001 mol/L * 20 mL = 0.001 mol/L * 0.02 L = 0.00002 moles
nCH3COOH = 1.5 mol/L *40 mL = 1.5mol/L * 0.04L = 0.06 moles
The reaction isL: HCl + CH3COOH <--------> Cl- + CH3COOH2+
HCl is a strong acid and acetic acid is a weak acid.
So, pH = -log[H+] due to HCl
pH = -log 0.00002 = 4.7
2.
nHCl = 0.001 mol/L * 20 mL = 0.001 mol/L * 0.02 L = 0.00002 moles
nCH3COONa = 2.5 mol/L *50 mL = 2.5mol/L * 0.05L = 0.125 moles
The reaction isL: HCl + CH3COONa <--------> NaCl + CH3COOH
Using Henderson-Hasselbalch equation,
pH = pKa + log [ base ] / [acid]
= -log Ka + log 0.125 - log 0.00002
= -log 1.75*10-5 + (-0.9 + 4.7)
= 4.76 + 3.8
= 8.6
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