Titration of HCl with CaCO3, then excess HCl titrated with NaOH

[i_love_science]

Homework Statement:

50.0 cm3 of 1.00 mol dm-3 hydrochloric acid solution, HCl(aq) was added to a piece of pure marble,
CaCO3 with a mass of 1.926 g. After all the marble had reacted the resulting solution was put into
a 100 cm3 volumetric flask and the volume made up to the mark with distilled water. 10.0 cm3 of
this solution was then titrated with 1.00 x 10-1 mol dm-3sodium hydroxide solution, NaOH(aq).
Calculate the volume of the sodium hydroxide solution required to neutralise exactly the excess
acid present in this 10.0 cm3 sample during this titration.

Relevant Equations:

titration

There is 0.019243 mol of CaCO3, and therefore 0.038485 mol of HCL. There is 0011505 mol of HCl in excess after reacting it with CaCO3. The concentration of HCl after water is added is 0.11515 mol/dm3. There is 0.0011515 mol of HCl in 10 cm3 of the new solution, and the same number of moles of NaOH reacted.

I got 0.0115 L of NaOH as my final answer, is it correct? Thanks.

 

[Borek]

Looks OK to me.