Question

How many grams of potassium hydrogen phthalate should

be weighed into a flask to standardizeNaOH if you wish

to use,of base for the titration?

Short answer

The quantity of potassium hydrogen phthalate (in grams) required to standardize NaOH in order to titrate 30 mL of the base solution is 0.31gms.

Step-by-step solution

Definition ofmolarity.

Molarity is one of the parameters used to express the concentration of a solution. It is expressed as,

\({\rm{Molarity = }}\frac{{{\rm{ number of moles of solute }}}}{{{\rm{ volume of solution in L}}}}\)

The moles of NaOH.

Molarity of NaOH is \(0.05{\rm{M}} = 0.05\;{\rm{mol}}/{\rm{L}}\).

The volume of\({\rm{NaOH}}\) to be titrated is 30 mL.

Therefore,

\(\begin{aligned}{}{\rm{moles of NaOH = Molarity \times Volume}}\\ = 0.05\;{\rm{mol}}/{\rm{L}} \times 30\;{\rm{mL}}\\{\rm{ = }}1.5 \times {10^{ - 3}}\;{\rm{mol}}\\ = 1.5{\rm{mmol}}\end{aligned}\)

 The molarity of the HCI. 

The number of moles of \({\rm{NaOH}}\) in \(0.05{\rm{M}}\) solution is equivalent to the number of moles of Potassium hydrogen phthalate \(({\rm{KHP}})\) required to standardize.

As,

\({\rm{moles = }}\frac{{{\rm{mass}}}}{{{\rm{molecular mass}}}}\)

Therefore, themass required in grams to standardize \(30\;{\rm{mL}}\) of \({\rm{0}}{\rm{.05M}}\) NaOH solution is

\(\begin{array}{c}{\rm{mass of KHP = moles KHP}} \times {\rm{molecular mass of KHP}}\\ = 1.5 \times {10^{ - 3}}\;{\rm{mol}} \times 204.22\;{\rm{g}}/{\rm{mol}}\\ = 0.31\;{\rm{g}}\end{array}\)