Question

The owner of a swimming pool supply company is offered what appears to be an exceptional deal on some concentrated hydrochloric acid (muriatic 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 the sample analyzed before agreeing to make a purchase. The chemist he hired determined the density of the solution to be 1.367g/mL. Then a 5.00mL aliquot of the acid was diluted to 250mL. A 50.0mL aliquot of the diluted sample requires 22.5mL of 0.300 M NaOH for titration to the phenolphthalein end point. Calculate the percent HCl in the original acid.

(answer is 18.0% HCl but need to know how to get it)

Answer 1

D = 1.367 g/mL of solution

V = 5 mL of HCl is used... then diluted to V = 250 mL

then, 50 mL are used...

NaOH + Hcl = H2O + NaCl

ratio is 1:1

mol of NaOH = mol of HCl

mol of NaOH = MV = 22.5*0.3 = 6.75 mmol of NaOH

mmol of HCl = 6.75 mmol

This is in 50 mL

[HCl] = mol/V = 6.75/50 = 0.135 M

this is fr the 50 mL

which was taken from the 250 mL

so

V = 250 mL; M = 0.135 M

this came from the 5 mL so:

M1*V1 = M2*V2

M2 = M1*V1/V2 = (0.135)*250/5 = 6.75 M

V = 5 mL; a M = 6.75 M

now, calculate % of HC

assume a basis of 1 Liter so

mol = 6.75 mol of HCl

1L solution --> mass = Density * Volume = 1.367 * 1000 mL = 1367 g of solution

mass of HCk = mol*MW = 6.75*36.5 = 246.375 g

mass of water= 1367 -246.375 = 1120.625 g of water

so...

% ww = mass of HCl / mass of solutin * 100% = 246.375 / 1367 *100 = 18.02 %

the soluton is actually 18%