Question

CAN SOMEONE HELP?✨

A student weighs out 1.118 g of impure KHP, dissolves the sample in deionized water, and titrates it with 0.1001 M NaOH solution. If the titration requires 27.10 mL NaOH solution, and none of the impurities react with NaOH, what is the percent KHP in the sample?

Answer 1

We can use the balanced chemical equation for the reaction between KHP (potassium hydrogen phthalate) and NaOH:

KHC8H4O4 + NaOH → NaKC8H4O4 + H2O

From the equation, we can see that 1 mole of KHP reacts with 1 mole of NaOH. We can use this information to calculate the moles of NaOH used in the titration:

moles of NaOH = concentration of NaOH x volume of NaOH used

moles of NaOH = 0.1001 mol/L x 0.02710 L

moles of NaOH = 0.002711 mol

Since 1 mole of NaOH reacts with 1 mole of KHP, the moles of KHP in the sample is also 0.002711 mol.

Now, we can use the mass of the sample to calculate the percent KHP in the sample:

percent KHP = (mass of KHP / mass of sample) x 100%

To find the mass of KHP in the sample, we need to first calculate the molecular weight of KHP:

KHC8H4O4: K = 39.10 g/mol, H = 1.01 g/mol, C = 12.01 g/mol, O = 16.00 g/mol

Molecular weight of KHP = 204.22 g/mol

mass of KHP = moles of KHP x molecular weight of KHP

mass of KHP = 0.002711 mol x 204.22 g/mol

mass of KHP = 0.5547 g

Now, we can calculate the percent KHP in the sample:

percent KHP = (mass of KHP / mass of sample) x 100%

mass of sample = 1.118 g

percent KHP = (0.5547 g / 1.118 g) x 100%

percent KHP ≈ 49.6%

Therefore, the percent KHP in the sample is approximately 49.6%.