Question

105 Chemistry I Laboratory Manual, 2017 Revision Questions: 1) What mass of KHP would you need in order to use 35.00 mL of 0.400 M NaOH to titrate it?

Answer 1

1)

Reaction is

KHP + NaOH $\rightarrow$ NaKP + H₂O.

mole ratio of KHP and NaOH = 1:1.

Volume of tbe solution = 35.00 mL = 0.035 L

Now, moles of NaOH = molarity × volume (L)

= 0.400× 0.035

= 0.014

Mass of KHP needed = moles of KHP × molar mass of KHP

= 0.014 (mol) × 204.2 (g/mol)

= 2.86 g.

2.

From the balanced reaction

Moles of KHP = Moles of NaOH

Volume of NaOH = 22.10 mL = 0.0221 L

Now, moles of NaOH = molarity × volume (L).

= 0.1101×0.0221

= 0.0024

So, moles of KHP = 0.0024

Now mass of KHP

= 0.0024× 204.2

= 0.490 g.

Now, percent of KHP in the sample

= ( mass of KHP ×100)/total mass)

= (0.490 × 100)/1.000)

= 49.0 %