CHEMISTRY

JAMB 2008 - Question 20

Chemistry 2008 JAMB Past Questions - Question 20: 0.05 mol dm³ HCI is neutralized by 25cm³ NaOH. If the volume of acid used is 32.00cm³ , what is the concentration of the base?

Choose the correct answers from the options given.
0.05 mol dm³ HCI is neutralized by 25cm³ NaOH. If the volume of acid used is 32.00cm³ , what is the concentration of the base?
A:
B:
C:
D:
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Correct Answer

D

Explanation

Certainly! To find the concentration of the base (NaOH) solution, we can follow these steps:

1. **Convert volumes to dm³:**
    - Acid volume: 32.00 cm³ * (1 dm³ / 1000 cm³) = 0.03200 dm³
    - Base volume: 25 cm³ * (1 dm³ / 1000 cm³) = 0.02500 dm³

2. **Calculate the moles of acid used:**
    - moles of acid = concentration of acid * volume of acid
    - moles of acid = 0.05 mol/dm³ * 0.03200 dm³ = 0.001600 mol

3. **Utilize the balanced chemical equation:**
    HCl (aq) + NaOH (aq) → NaCl (aq) + H₂O (l)
    Since the acid and base react in a 1:1 ratio according to the equation, the moles of base used are equal to the moles of acid neutralized.

4. **Calculate the concentration of the base:**
    - concentration of base = moles of base / volume of base
    - concentration of base = 0.001600 mol / 0.02500 dm³ ≈ 0.0640 mol/dm³

Therefore, the concentration of the base (NaOH) solution is approximately **0.0640 mol/dm³**.

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