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*2025 WAEC chemistry practical alternative A theoretical value for VA*

From the instruction paper

Volume of concentrated HCl (V₁) = 8.6cm³

Mass concentration of NaHCO₃ = 8g/dm³

My workings

% purity of HCl = 36

Specific gravity of HCl = 1.18

Molar mass of NaHCO₃ = (23) + (1) + (12) + (48) = 84g/mol

Molar mass of HCl = (1) + (35.5) = 36.5g/mol

VB = 25.0cm³

Equation of the reaction
HCl + NaHCO₃ -----> NaCl + H₂O + CO₂

nA = 1 and nB = 1

To get molar concentration (C₁) of the concentrated acid

Formula to use is

C₁ = (10 × % purity × specific gravity)/Molar mass

C₁ = (10 × 36 × 1.18)/36.5
C₁ = 424.8/36.5
C₁ = 11.6383
C₁ = 11.6mol/dm³
This C₁ is the molar concentration of concentrated HCl

To get C₂ which is molar concentration of the dilute HCl

Formula to use is
C₁V₁= C₂V₂
(11.6 × 8.6cm³) = (C₂ × 1000cm³ for dilution)

99.76 = 1000C₂
Divide both sides by 1000
C₂ = 0.09976
C₂ = 0.0998mol/dm³
This C₂ is our CA and it is the molar concentration of the dilute HCl

To get molar concentration of NaHCO₃ (i.e CB)

Formula to use is
CB = (Mass concentration of B)/(Molar mass of B)

CB = 8/84
CB = 0.09523 to 3 s.f
CB = 0.0952mol/dm³

To get VA

Formula to use is
VA = (CBVBnA)/(CAnB)

VA = (0.0952 × 25 × 1)/(0.0998 × 1)
VA = 2.38/0.0998
VA = 23.8476 to 3 s.f
VA = 23.80cm³

*2025 WAEC chemistry practical alternative A theoretical value for VA = 23.80cm³*

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