25.00cm^3 of sodium trioxocarbonate (IV) solution neutralizes 22.50cm^3 of 0.05moldm^-3 hydrochloric acid. Calculate the concentration of Na2CO3.

25.00cm^3 of sodium trioxocarbonate (IV) solution neutralizes 22.50cm^3 of 0.05moldm^-3 hydrochloric acid. Calculate the concentration of Na2CO3.

A. 0.023

B. 0.046

C. 0.230

D. 0.460

Calculation:

CA VA}/{CB VB} = nA/nB

Where:

• (CA, VA) → concentration & volume of acid

• (CB, VB) → concentration & volume of base

• (nA:nB) → ratio from the balanced equation

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Step 1: Write the balanced equation

Na2CO3 + 2HCl一> 2NaCl + H2O + CO2

• Ratio (nA:nB = 1:2)
  (1 mole Na₂CO₃ reacts with 2 moles HCl)

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Step 2: Apply the formula

{{C{Na₂CO₃} * V{Na₂CO₃}}/{C{HCl} *V{HCl}} = {1}/{2}

Plug in the values:

{C{Na₂CO₃} * 25.00}/{0.05 *22.50} = {1}/{2}

• Convert cm³ to same units (they cancel if both in cm³).

C{Na₂CO₃} * 25 = 0.05 * 22.5 * {1}/{2}

C{Na₂CO₃} * 25 = 0.5625
C{Na₂CO₃} = {0.5625}/{25} = 0.0225 approx 0.023 mol/dm³

✅ Same answer: 0.023 mol/dm³

PQN: 31