Question

# a. A student dissolves 1.1007 grams of their KHP unknown in 50.0 mL of water. The...

a. A student dissolves 1.1007 grams of their KHP unknown in 50.0 mL of water. The molarity of the student’s NaOH titrant is 0.0999 M. If it takes 40.25 mL of the sodium hydroxide titrant to reach the equivalence point, what is the wt% KHP (MW=204.23) in the unknown? b. If the student uses 55.0 mL of water instead of 50.0 mL to dissolve the unknown, what affect, if any, does this have on the calculated weight % of the unknown. Explain your answer in one complete sentence.

m = 1.1007 g of sample of KHP

V = 50 mL

[NaOH] = 0.0999 M

if

V base = 40.25 mL required

NaOH + KHP = KNaP + H2O

ratio is 1:1

mmol of NaOH = M*V = 0.0999*40.25 = 4.020975 mmol of 4.020975*10^-3 mol

now..

mol of KHP = 4.020975 mmol or 4.020975*10^-3 mol

mass = mol*MW = (4.020975*10^-3) * 204.23 = 0.8212 g

%mas sof KHP = mass of KHP / Total mas sample * 100%

% KHP = 0.8212 / 1.1007 * 100 = 74.607 %

b)

V = 55 mL of water instead of 50 mL....

what effect will have on %

NONE, since we are talking about moles equivalents

if all mass is sitll dissolving (1.1007 g) in the 55 mL, then the 1.1007 g of sample are present

so they will still have the SAME content of KHP

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