Question

You are given solutions of hcl and naoh and must determine their concentrations. you use 27.5 ml of naoh to titrate 100.0 ml of hcl and 18.4 ml of naoh to titrate 50.0 ml of 0.0782 m h2so4. based on this data, what were the concentrations (molarities) of the hcl and naoh solutions, respectively

Answer 1

1) Start by standardizing the solution of NaOH by using the solution of H2SO4 whose concentration is known.

2) Equation:

2Na OH + H2SO4 --> Na2 SO4 + 2H2O

3) molar ratios

2 mol NaOH : 1 mol H2SO4

4) Number of moles of H2SO4 in 50.0 ml of 0.0782 M solution

M = n / V => n = M*V = 0.0782 M * 0.050 l = 0.00391 mol H2SO4

5) Number of moles of NaOH

2 moles NaOH / 1mol H2SO4 * 0.00391 mol H2SO4 = 0.00782 mol NaOH

6) Concentration of the solution of NaOH

M = n / V = 0.00782 mol / 0.0184 ml = 0.425 M

7) Standardize the solution of HCl

Chemical reaction:

NaOH + HCl --> NaCl + H2O

8) Molar ratios

1 mol NaOH : 1 mol HCl

9) Number of moles of NaOH in 27.5 ml

M = n / V => n = M * V = 0.425 M * 0.0275 l = 0.01169 moles NaOH

10) Number of moles of HCl

1 mol HCl / 1mol NaOH * 0.01169 mol NaOH = 0.01169 mol HCl

11) Concentration of the solution of HCl

M = n / V = 0.01169 mol / 0.100 l = 0.1169 M

Rounded to 3 significant figures = 0.117 M

Answers:

[NaOH] = 0.425 M
[HCl] = 0.117 M