Using the laboratory procedure, a student determined the percent KHP in an impur

Question

Using the laboratory procedure, a student determined the percent KHP in an impure sample of KHP. A 2.740 g sample of impure KHP required 39.50 mL of .1255 mL of 0.1255 M NaOH solution for titration. Calculate the number of moles of NaOH required for the titration. Calculate the number of moles of KHP present in the impure sample of KHP. Calculate the number of grams of KHP present in the impure sample. Calculate the percent KHP in the impure sample. Sodium carbonate is a reagent that may be used to standardize acids in the same way as KH was in this experiment. It was found that a 0.4981 g sample of sodium carbonate required 25.72 mL of a sulfuric acid solution to reach the end point for the reaction. What is the molarity of the H_2SO_4?

Explanation / Answer

m = 2.74 g of KHP

V = 39.50 mL of M = 0.1255 M of NaOH

ratio is 1:1

NaOH + KHP = H2O + KNaP

so

mmol of NaOH = MV = 39.50*0.1255 = 4.95725 mmol

a) moles of NAOH = 4.95725 *10^-3 mol = 0.0049572 mol of NaOH

then

mmol of KHP = 4.95725 mmol

b) mmol of KHP = 4.95725 mmol

mol of KHP =  0.0049572 mol of KHP

so..

mass = mol*MW = (4.95725*10^-3)(204.22 ) = 1.0123 grams of KHP

d)

%KHP = mass of KHP / total mass * 100 = 1.0123 /2.74 * 100 = 0.36945*100 = 36.9 %

Q2

mol of Na2CO3 = mass/MW = 0.4981/105.9888 = 0.00469 mol

so...

ratio is 1:1

mol of H2SO4 = 0.00469mmol

that is

M = mol/V = (0.00469)/(25.72*10^-3) = 0.1823 M of H2SO4

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