Question

How much 0.0135 M HCl is required to titrate 235 mL of 0.0135 M Na2CO3 (to the first endpoint) or 235 mL of 0.0135 MNaHCO3? 1st attempt See Periodic Table Part 1 (0.5 point) To titrate Na2CO3 (first endpoint)? mL HCI Part 2 (0.5 point) To titrate NaHCO3? mL HCI

Answer 1

Let us write the balanced equation for first step:
Na₂CO₃ + HCl ====> NaHCO₃ + NaCl
Reaction type: double replacement

Volume of Na₂CO₃ = 235 ml

Concentration of Na₂CO₃ = 0.0135 M

Moles of Na₂CO₃ = 235 x 0.0135 / 1000 =  0.0031725‬ Moles

Moles of HCl required to react =   0.0031725‬ Moles

Volume of HCl required =   0.0031725 x 1000 / 0.0135 = 235 ml

Similarly it required another 235 ml of HCl required to react with NaHCO3 .

Part 1 (0.5 point) To titrate Na2CO3 (first endpoint)? 235 mL HCI Part 2 (0.5 point) To titrate NaHCO32 235 mL HCI