# Backtitration Problem

1. Sep 28, 2014

### orchidee7

• Please post homework and homework like problems in the homework forum
I'm stuck on how to find how many ml of NaOH were used in a back titration.

A 0.6000 g sample of K2CO3 (138.2055 g/mol) is dissolved in enough water to make 200.0 mL of solution A. A 20.00 mL aliquot of solution A is taken and put into an Erlenmeyer flask. To the flask is added 20.00 mL of 0.1700 M HCl:

K2CO3(aq) + 2HCl(aq)2KCl(aq) + H2O(l) + CO2(g)

The resulting solution is then titrated with 0.1048 M NaOH.

NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq)

How many mL of NaOH are used? (24.16)

I have:
0.6000gK2CO3 x (1/138.2055gK2CO3) x (1/.200ml) = 0.021707 M K2CO3.
0.170M HCl x (1/0.04ml) = 0.0034 mol HCl and 0.0034 mol NaOH (1:1 ratio)

From here I don't know what to do to find the ml of NaOH required to solve the equation. Could anybody show me how or give me some hints? Thank you.

2. Sep 29, 2014

### Staff: Mentor

What was the amount of the excess HCl (first reaction)?