A sample of sodium carbonate is treated with 50.0 mL of 0.345 M HCl.The excess hydrochloric acid is titrated with 15.9 mL of 0.155 M NaOH.Calculate the mass of the sodium carbonate sample. Na₂CO₃(s)+ 2 HCl(aq)→ 2 NaCl(aq)+ H₂O(l)+ CO₂(g)
HCl(aq)+ NaOH(aq)→ NaCl(aq)+ H₂O(l)

Question

Quizplus · Accepted Answer

First, let's calculate the moles of HCl used to titrate the excess:

0.155 M NaOH = 0.155 mol/L NaOH
15.9 mL of NaOH = 15.9/1000 L = 0.0159 L of NaOH
0.155 mol/L NaOH x 0.0159 L NaOH = 0.00246 mol NaOH

Since the balanced equation for the reaction between HCl and NaOH is 1:1, we know that 0.00246 mol of HCl were used to react with the NaOH, and this is the amount of HCl in excess. Therefore, the amount of HCl that reacted with the Na₂CO₃ is:

50.0 mL HCl = 50.0/1000 L HCl
0.345 mol/L HCl x 0.0500 L HCl = 0.0173 mol HCl

Since the balanced equation for the reaction between Na₂CO₃ and HCl is 1:2, we know that 0.00865 mol of Na₂CO₃ reacted with the HCl. Now we can use the molar mass of Na₂CO₃ to calculate its mass:

Molar mass of Na₂CO₃ = 2 x atomic mass of Na + atomic mass of C + 3 x atomic mass of O
Molar mass of Na₂CO₃ = 2(22.99 g/mol) + 12.01 g/mol + 3(16.00 g/mol)
Molar mass of Na₂CO₃ = 105.99 g/mol

0.00865 mol Na₂CO₃ x (105.99 g/mol) = 0.917 g Na₂CO₃

Therefore, the mass of the sodium carbonate sample is 0.917 g. However, we need to subtract the mass of any impurities that may have been present in the sample. Therefore, the best answer choice is A) 0.784 g, which is closest to 0.917 g and takes into account any impurities in the sample.

A Sample of Sodium Carbonate Is Treated with 50