You titrate a 0.53 g sample of KHP with NaOH and find that it takes 23.38 mL of

Question

You titrate a 0.53 g sample of KHP with NaOH and find that it takes 23.38 mL of base to reach the endpoint. Calculate the concentration in molarity of the NaOH solution. Do not use scientific notation to report your answer.

Accepted characters: numbers, decimal point markers (period or comma), sign indicators (-), spaces (e.g., as thousands separator, 5 000), "E" or "e" (used in scientific notation). Complex numbers should be in the form (a + bi) where "a" and "b" need to have explicitly stated values. For example: {1+1i} is valid whereas {1+i} is not. {0+9i} is valid whereas {9i} is not. You titrate a 0.53 g sample of KHP with NaOH and find that it takes 23.38 mL of base to reach the endpoint. Calculate the concentration in molarity of the NaOH solution. Do not use scientific notation to report your answer. Accepted characters: numbers, decimal point markers (period or comma), sign indicators (-), spaces (e.g., as thousands separator, 5 000), "E" or "e" (used in scientific notation). Complex numbers should be in the form (a + bi) where "a" and "b" need to have explicitly stated values. For example: {1+1i} is valid whereas {1+i} is not. {0+9i} is valid whereas {9i} is not. You prepare a vinegar sample by diluting 5.00 mL of vinegar with 20.00 mL of water. You titrate the resulting solution with 0.13 M NaOH and find that it requires 29.02 mL of base to reach the end point. Calculate the concentration in molarity of acetic acid in the vinegar.

Explanation / Answer

(1) NaOH + KHP => NaKP + H2O

Moles of NaOH = moles of KHP = mass/molar mass of KHP

= 0.53/204.22 = 0.0025952 mol


Molarity = moles/volume of NaOH

= 0.0025952/0.02338

= 0.111 M (or approximately 0.11 M)


(2) NaOH + CH3COOH => CH3COONa + H2O

Moles of CH3COOH = moles of NaOH = volume x concentration of NaOH

= 29.02/1000 x 0.13 = 0.0037726 mol


Molarity = moles of CH3COOH/volume of vinegar

= 0.0037726/0.00500

= 0.7545 M (or approximately 0.75 M)

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