A 0.205 g sample of CaCO3 (Mr = 100.1 g/mol) is added to a flask along with 7.50 mL of 2.00 M HCl. CaCO3(aq) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g) Enough water is then added to make a 125.0 mL solution. A 10.00 mL aliquot of this solution is taken and titrated with 0.058 M NaOH. NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq) How many mL of NaOH are used?

Question

contestada

Respuesta :

pstnonsonjoku pstnonsonjoku · Answer 1 · 2022-03-29T12:41:47+02:00

The required volume of sodium hydroxide is 15 mL.

The term concentration refers to the amount of substance present. It is the quotient of the number of moles and volume of solution.

We have the reaction; 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)

Number of moles of CaCO3 = 0.205 g / 100.1 g/mol = 0.0020 moles

Number of moles of HCl = 2.00 M * 7.50/1000 = 0.015 moles

Since 2 moles of HCl reacts with 1 mole of CaCO3

x moles of HCl reacts with 0.0020 moles

x = 0.004 moles of HCl

This means that HCl is in excess by the amount 0.011 moles.

Now consider the reaction; NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq)

Concentration of HCl = 0.011 moles/0.125L = 0.088 M

From;

CAVA/CBVB = na/nb

CAVAnb = CBVBna

VB = CAVAnb/CBna

VB = 0.088 M * 10mL * 1/0.058 M * 1

VB = 15 mL

Learn more about concentration: https://brainly.com/question/3045247