Question

After doing multiple titrations, your NaOH solution is determined to have a mean concentration value of 0.100 M. Given you are to assume your unknown acid is 75.0% KHP, how many grams of your unknown will you need to use 15.00 mL of your 0.100 M standardized NaOH

Answer 1

Answer: The mass of unknown acid needed is 0.230 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaOH solution = 0.100 M

Volume of solution = 15.00 mL = 0.015 L    (Conversion factor:  1 L = 1000 mL)

Putting values in above equation, we get:

$0.100M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.100mol/L\times 0.015L)=0.0015mol$

The chemical reaction for the reaction of KHP and NaOH follows

$KHC_8H_4O_4(aq.)+NaOH\rightarrow KNaC_8H_4O_4(aq.)+H_2O(l)$

By Stoichiometry of the reaction:

1 mole of NaOH reacts with 1 mole of KHP.

So, 0.0015 moles of NaOH will react with = $\frac{1}{1}\times 0.0015=0.0015mol$ of KHP

• To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of KHP = 0.0015 moles

Molar mass of KHP = 204.22 g/mol

Putting values in above equation, we get:

$0.0015mol=\frac{\text{Mass of KHP}}{204.22g/mol}\\\\\text{Mass of KHP}=(0.0015mol\times 204.22g/mol)=0.306g$

We are given:

Mass of unknown acid = 75 % of Mass of KHP

So, mass of unknown acid = $\frac{75}{100}\times 0.306=0.230g$

Hence, the mass of unknown acid needed is 0.230 grams