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Determine the molarity of a solution prepared by dissolving 0.198 g of "KHP" (C8H5O4K) in enough water to make 125 mL of solution.
When the solution is found and titrated with aqueous sodium hydroxide, the endpoint is reached after the addition of 27.80 mL of base solution. How many moles of NaOH reacted? What is the concentration of the NaOH?

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Answer:

0.007756M KHP; 0.0009695 moles NaOH reacts; 0.03488M NaOH.

Explanation:

Potassium hydrogen phthalate, KHP, is a salt used as standard to determine concentration of basic solutions as NaOH solutions.

To find molarity of the KHP solution we need to convert mass of KHP to moles and divide this in 125mL = 0.125L:

Moles KHP -Molar mass: 204.22g/mol-:

0.198g * (1mol / 204.22g) = 0.0009695 moles KHP

Molarity:

0.0009695 moles KHP / 0.125L = 0.007756M KHP

NaOH reacts with KHP as follows:

NaOH + KHP → Na⁺ + KP⁻ + H₂O

That means 1 mole of NaOH reacts per mole of KHP, that means moles of NaOH that reacts are = Moles of KHP added:

0.0009695 moles NaOH reacts

In 27.80mL = 0.02780L the NaOH was contained. Molarity is:

0.0009695 moles NaOH / 0.02780L =

0.03488M NaOH

Cricetus

Moles of NaOH added will be "0.034".

According to the question,

→ [tex]Molarity = \frac{Number \ of \ moles}{Volume}[/tex]

By substituting the values, we get

                  [tex]= \frac{0.198}{204}\times \frac{1000}{125}[/tex]

                  [tex]= 0.0077 \ M[/tex]

Let,

  • The number of moles added be "[tex]M_2[/tex]".

Now,

→            [tex]M_1 V_1 = M_2 V_2[/tex]

→ [tex]0.0077\times 125=M_2\times 27.8[/tex]

→                [tex]M_2 = 0.034 \ moles[/tex]

Thus the above answer is appropriate.

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