Dissolving 4g impure NaOH in water and complete the volume to 200 ml. If 10 of this solution neutralizes with 15 ml of 0.2 M Hcl.Calculate the percentage of the NaOH in the sample

Answer 1

Let us first write the balanced equation for the neutralization reaction:

HCl + NaOH ==> NaCl + H2O .. balaneced equation

Next, we will calculate the MOLES of HCl that are present:

15 ml x 1 L / 1000 ml x 0.2 mol / L = 0.003 mols HCl

Then, calculate MOLES of NaOH that was used for neutralization:

0.003 mol HCl x 1 mol NaOH / mol HCl = 0.003 mols NaOH

Next, convert moles NaOH to grams NaOH using the molar mass:

0.003 mols NaOH x 40 g / mole = 0.12 g NaOH present

At this point, we have calculated the mass of NaOH present in 10 ml of the original 200 mls of sample. So we can now calculate the mass of NaOH in the original 200 mls:

0.12 g NaOH / 10 ml x 200 ml = 2.4 g NaOH in the original sample

% NaOH = mass of NaOH / mass of sample (x100%)

%NaOH = 2.4 g / 4 g (x100%) = 60% NaOH