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?

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

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

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 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

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