Question

A 0.125L solution of NaOH is neutralized by 0.0425L of a 0.65 M H2SO4 solution. What is the concentration of the NaOH solution?

Answer 1

The concentration of the NaOH solution, calculated using the stoichiometry of the neutralization reaction with H2SO4, is found to be 0.442 M.

Step-by-step explanation:

To calculate the concentration of the NaOH solution, we need to use the stoichiometry of the neutralization reaction that occurs between NaOH and H2SO4. The reaction is as follows:

• H2SO4 (aq) + 2NaOH (aq) → Na2SO4 (aq) + 2H2O (l)

First, calculate the moles of H2SO4 that reacted by using its concentration and volume:

• Moles H2SO4 = 0.65 M × 0.0425 L = 0.027625 mol

The reaction shows that 1 mole of H2SO4 reacts with 2 moles of NaOH, so:

• Moles NaOH = 2 × Moles H2SO4 = 2 × 0.027625 mol = 0.05525 mol

Now, use the moles of NaOH and the volume of the NaOH solution to find its concentration:

• Concentration of NaOH = Moles NaOH / Volume NaOH = 0.05525 mol / 0.125 L = 0.442 M

The concentration of the NaOH solution is therefore 0.442 M.

Answer 2

0.442 M

Step-by-step explanation:

The equation of the reaction must first be written down before the problem can be correctly solved.

2NaOH(aq) + H2SO4(aq) ------> Na2SO4(aq) + 2H2O(l)

Given that;

Concentration of acid CA= 0.65 M

Concentration of base CB= ????

Volume of acid CA= 0.0425L

Volume of base CB= 0.125L

Number of moles of acid = 1

Number of moles of base = 2

From;

CAVA/CBVB= NA/NB

CAVANB= CBVBNA

CB= CAVANB/VBNA

Substituting values;

CB= 0.65 × 0.0425 × 2 / 0.125 × 1

CB= 0.442 M