Answer to Question #96167 in General Chemistry for Brittany Wallace

Question

KHC8H4O4 (aq) + NaOH (aq) → KNaC8H4O4 (aq) + H2O(l)
KHP is an abbreviation for KHC8H4O4 and its molar mass is 204.2 g/mol.

1. What is the molarity of a NaOH solution if 26.50 mL is required to neutralize 0.4150 g of pure KHP?








2. What is the percentage of KHP in an impure sample of KHP that weighs 0.4200 g and requires 14.00 mL of 0.098 M NaOH to neutralize it?

Answer 1

Q96167

Solution:

1. As the given equation is already balanced, using mole-mole analysis, we get:

$moles of KHP reacted = moles of NaOH reacted$ -----(x)

$moles = given weight/molecular weight$

thus, moles of KHP reacted = $0.4150g / 204.2g$

= 0.002 mol.

thus, $molarity = no. of moles used/volume(in litres)$ (for NaOH)

= $(0.002/26.5)*1000$

= 0.075 M (Answer)

2. $moles = Molarity * Volume used (in litres)$

=$0.098*14/1000$

=1.372 milimoles.

hence, moles of NaOH used = 1.372 milimoles

hence, moles of KHP used = 1.372 milimoles (from (x))

$weight = moles*molecular weight$

Hence, weight of KHP used = $1.372*204.2/1000$

= 0.28 gms.

$purity=weight used/weight taken*100$

Thus, purity of KHP sample = $(0.28/0.42)*100$

= 66.67% (Answer)