Question

In a standardization titration involving hydrochloric acid and sodium carbonate, a student recoirded the following results for the volume of hydrochloric acid used against 10.00mL of the sodium carbonate solution : 15.60; 14.50; 14.70 aqnd 14.20. If the concentration of the
Na2CO3
standard solution is 0.75moldm^-3, calculate the concentration of the HCl solution
1.02 Ml
1.03mL
1.07mL
1.04M

Answer 1

Answer on Question #51888, Chemistry, Other

Question: In a standardization titration involving hydrochloric acid and sodium carbonate, a student recorded the following results for the volume of hydrochloric acid used against 10.00 mL of the sodium carbonate solution: 15.60; 14.50; 14.70 and 14.20. If the concentration of the Na₂CO₃ standard solution is 0.75 mold/m^3, calculate the concentration of the HCl solution

1.02 Ml

1.03 mL

1.07 mL

1.04 M

Answer: The average titre value for HCl is: $(15.60 + 14.50 + 14.70 + 14.20) / 4 = 14.75 \text{mL}$

Then we need to use the formula: $M_{\text{acid}} V_{\text{acid}} = 2 M_{\text{base}} V_{\text{base}}$; $M_{\text{acid}} = M_{\text{base}} V_{\text{base}} / V_{\text{acid}}$; $M_{\text{acid}} = (2 \times 0.75 \times 0.01 \text{L}) / 0.01475 = 1.02 \text{M}$

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