Question

The concentration of a certain sodium hydroxide solution was determined by using the solution to titrate a sample of potassium hydrogen phthalate (abbreviated as KHP). KHP is an acid with one acidic hydrogen and a molar mass of 204.22 g/mol. In the titration, 58.98 mL of the sodium hydroxide solution was required to react with 0.4000 g KHP. Calculate the molarity of the sodium hydroxide.

Answer 1

Answer:

The molarity of the sodium hydroxide is 0.0332 M

Explanation:

KHP contains only one hydrogen, one mole of KHC₈H₄O₄ can be titrated with one mole of NaOH.

No. of moles of KHC₈H₄O₄ = mass / molar mass = 0.4000 / 204.22

= 1.96 x 10⁻³ moles

∴ 1.96 x 10⁻³ moles of KPH can titrate 1.96 x 10⁻³ moles of NaOH

conc of NaOH = moles x (1/volume)

                        = 1.96 x 10⁻³ (1/0.05898)

       [NaOH]    = 0.0332 M

Answer 2

Answer: The molarity of NaOH is 0.033 M

Explanation:

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

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

Given mass of KHP = 0.400 g

Molar mass of KHP = 204.22 g/mol

Putting values in above equation, we get:

$\text{Moles of KHP}=\frac{0.40g}{204.22g/mol}=0.00195mol$

The chemical reaction for the reaction of KHP and NaOH follows  :

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

1 mole of KHP reacts with 1 mole of NaOH

So, 0.00195 moles of KHP will react with =  of NaOH

To calculate the molarity, we use the equation:

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

Moles of NaOH solution = 0.00195 moles

Volume of solution = 58.98 mL = 0.05898 L    (Conversion factor:  1 L = 1000 mL)

Putting values in above equation, we get:

$\text{Molarity of NaOH}=\frac{0.00195}{0.05898L}\\\\\text{Molarity of NaOH}=0.033M$

Hence, the molarity of NaOH is 0.033 M