Question

If a 0.4856 gram sample of khp is dissolved in sufficient water to prepare 250 ml of solution, and 25 ml of the solution requires 18.76ml of sodium hydroxide solution to reach the equivalence point, what is the molarity of the sodium hydroxide?

Answer 1

Mass of potassium hydrogen pthalate KHP is 0.4856 g, its molar mass is 204.22 g/mol, number of moles of KHP can be calculated as follows:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass, putting the values,

$n=\frac{0.4856 g}{204.22 g/mol}=0.00237 mol$

This will be number of moles of NaOH at equivalent point.

Detailed calculations:

Molarity is defined as number of moles in 1 L of solution, for 250 mL of solution, molarity will be:

$M=\frac{0.00237 mol}{250 \times 10^{-3}L}=0.009511 M$

For 25 mL, apply dilution law as follows:

$M_{1}V_{1}=M_{2}V_{2}$

Putting the values,

$0.009511\times 250=M_{2}\times 25 mL$

On rearranging,

$M_{2}=\frac{0.009511\times 250}{25}=0.09511 M$

Convert molarity into number of moles,

$n=M\times V=0.09511 mol/L\times 25\times 10^{-3}L=0.00237 mol$

At equivalent point, number of moles of KHP will be equal to NaOH, thus, number of moles of NaOH will be 0.00237 mol.

Calculation for molarity:

Volume of NaOH is 18.75 mL, thus, molarity can be calculated as follows:

$M=\frac{n}{V}$

Putting the values,

$M=\frac{0.00237 mol}{18.75\times 10^{-3}L}=0.1264 M$

Therefore, molarity of NaOH is 0.1264 M