Question

Determine the molarity of NaOH to three decimal places if 21.77 mL of this NaOH were used to titrate 0.305 g of KHP (MM 204.22 g/mol)

Answer 1

Answer

The molarity of NaOH to three decimal places is 0.069 mol/L

Explanation

Given:

Volume of NaOH used = 21.77 mL

Reacting mass of KHP = 0.305 g

Molecular Mass of KHP = 204.22 g/mol

What to find:

The molarity of NaOH to three decimal places.

Step-by-step solution:

The first step is to write the balanced stoichiometric chemical equation of the reaction:

$HKC_8H_4O_4+NaOH\rightarrow NaKC_8H_4O_4+H_2O$

From the balanced chemical equation;

1 mol KHP reacts with 1 mol NaOH

Thus, moles of KHP is:

$\text{Moles of KHP }=\frac{Reacting\text{ mass}}{MM\text{ of KHP}}=\frac{0.305\text{ g}}{204.22\text{ g/mol}}=1.493\times10^{-3}\text{ mol}$

Since, 1 mol KHP reacts with 1 mol NaOH from the balanced equation,

Therefore, 1.493 x 10⁻³ mol KHP will react with:

$\frac{1\text{ mol NaOH }\times1.493\times10^{-3}mol\text{ KHP}}{1\text{ mol KHP}}=1.493\times10^{-3}mol\text{ NaOH}$

The last step is to calculate the molarity of NaOH:

Conversion factor:

1 mL = 10⁻³ L

21.77 mL = 2.177 x 10⁻³ L

Molarity of NaOH is:

$\begin{gathered} \text{Molarity of NaOH }=\frac{Moles\text{ of NaOH}}{Volume\text{ of NaOH in L}}=\frac{1.493\times10^{-3}\text{ mol}}{2.177\times10^{-3}\text{ L}} \\ \text{Molarity of NaOH }=0.069\text{ mol/L} \end{gathered}$