Calculate the number of moles of OH- added during titration. Concentration of Ac

Question

Calculate the number of moles of OH- added during titration.
Concentration of Acid (HCl): .104M Concentration of Base (NaOH): .115M Mass of KHP used in titration: .121g Total Volume used in titration: 5.25mL
Total Volume is made up of KHP dissolved in water. The lab is titrating the KHP and water solution to neutrality (phenolphthaelin is the indicator) by the addition of the standardized base NaOH. The amount of water is inconsequential. Calculate the number of moles of OH- added during titration.
Concentration of Acid (HCl): .104M Concentration of Base (NaOH): .115M Mass of KHP used in titration: .121g Total Volume used in titration: 5.25mL
Total Volume is made up of KHP dissolved in water. The lab is titrating the KHP and water solution to neutrality (phenolphthaelin is the indicator) by the addition of the standardized base NaOH. The amount of water is inconsequential.
Concentration of Acid (HCl): .104M Concentration of Base (NaOH): .115M Mass of KHP used in titration: .121g Total Volume used in titration: 5.25mL
Total Volume is made up of KHP dissolved in water. The lab is titrating the KHP and water solution to neutrality (phenolphthaelin is the indicator) by the addition of the standardized base NaOH. The amount of water is inconsequential.

Explanation / Answer

moles of KHP = mass / molar mass

= 0.121 / 204.22

= 5.92 x 10^-4

KHP + NaOH --------------------------> NaKP + H2O

moles of KHP = moles of NaOH

5.92 x 10^-4 = moles of NaOH

so moles of OH- during titration = 5.92 x 10^-4

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