Question

calculate how many mL of a 0.32 M solution of HCl must be added to an aqueous solution containing 4 g of Na2CO3 to obtain a solution at pH = 10. H2CO3 Ka1 = 4,5x10 -7 Ka2 = 4.8x 10 -10

A. 99,4

B. 80,8

C. 87,5

D. 33,9

Answer 1

NOTICE THAT YOU HAVE A WRONG Ka2 VALUE!!!! Ka2 = 4.8x10^-11 instead of  Ka2 = 4.8x 10 -10

A solution of Na₂CO₃ which HCl was added, will contain CO₃^2- and HCO₃^- in similar concentrations.

pH = pKa + Log ( [A-] /[HA] )

10 = 10.32 + Log ( [CO₃^2- ] / [HCO₃^- ] )

[CO₃^2- ] / [HCO₃^- ] = 0.48 ----> desired ratio of acid and base

On the other hand...

4g of Na₂CO₃ = 4g/106g*mol^-1 = 0.0377 moles of Na₂CO₃

Let's assume HCl reacts completely with Na₂CO₃ according to:

HCl + Na₂CO₃ -------> 2Na⁺ + Cl^- + HCO₃^- , so almost all the HCO₃^-? formed comes from that reaction.

Then: [CO₃^2- ] / [HCO₃^- ] = 0.48 and also [CO₃^2- ] +  [HCO₃^- ] = 0.0377 moles

0.0377-[HCO₃^- ] = 0.48 *  [HCO₃^- ] ---------> [HCO₃^- ] = 0.0254moles ;  [CO₃^2- ] = 0.0122 moles

Now, to generate  [HCO₃^- ] = 0.0254moles we need 0.0254moles of HCl

0.32 moles of HCl -------> 1000 ml of HCl solution

0.0254 moles of HCl ----> 80 ml of HCl solution