Question

The owner of a swimming pool supply company is offered what appears to be an exceptionally good deal on some concentrated muriatic acid (hydrochloric acid). The solution is supposed to be 31.25% HCl by weight. The store owner has some suspicions about the seller and decides to have a sample analyzed before agreeing to make a purchase. The chemist he hired determined the density of the solution to be 1.367 g/ml. Then a 5.0 ml aliquot of the acid was diluted to 250 ml. A 50.0 ml aliquot of the diluted sample required 22.5 ml of 0.30 M NaOH for titration to the phenolphthalein end point. Calculate the percentage of HCl in the original acid and tell the owner what he should do

Answer 1

22.5 ml of 0.30 M NaOH was required to titrate 50 ml of the diluted HCl solution.

From the equation V₁S₁ = V₂S₂, we can calculate strength of the diluted HCL solution. Let the strength be S₂.

Therefore, 22.5 ml x 0.30M = 50 ml x  S₂

S₂ = (22.5 x 0.30)/ 50 = 0.135 M

This solution was obtained by 50 times dilution of original solution (5 ml diluted to 250 ml).

So, original strength was = (0.135 x 50) M = 6.75 M

Molar mass of HCl = 36.46 g/mol

Therefore, in 1000 ml of original acid there is = (6.75 x 36.46) g HCl = 246.11 g

Thus, 100 ml solution contains = 246.11/10g = 24.61 g

In other words, the HCl solution is 24.61 % by weight instead of 31.25 %. The acid solution is dilute and the owner should not take it.