Answer to Question #148511 in General Chemistry for DanYale WIlson

Question

volume of Vinegar, mL 10.00 10.00 10.00



final NaOH volume, mL 25.20 25.35 25.08

initial NaOH volume, mL 0.00 0.00 0.00

vol. NaOH used, mL ________ ________ ________

Calculate Moles of NaOH using the Molarity of NaOH from the previous question

Moles of NaOH used ________ ________ ________

moles of vinegar neutralized ________ ________ ________

Answer 1

Q148511

We will need molarity from the first part. So we will first find the molarity of NaOH and then use it in the

second part of the problem.

Standardizing NaOH with KHP (molar mass 204.2 g/mole)

mass of flask + KHP = 99.829 g

mass of flask = 98.326 g

Mass of KHP = __________ g

moles of KHP _____________

initial buret reading 0.05 mL

final buret reading 25.35 mL

Volume of NaOH, mL ____________

moles of NaOH needed neutralize KHP _______________

Molarity of NaOH? _______________

Solution :

1) mass of KHP = mass of flask + KHP - mass if flask

= 99.829g – 98.326g

= 1.503 g.

2) moles of KHP = 1.503 g KHP * 1 mol KHP/204.2 g KHP = 0.00736 mol KHP.

3) Volume of NaOH, mL = Final burette reading – Initial burette reading

= 25.35 mL - 0.05 mL

= 25.30 mL .

4) For finding the moles of NaOH needed to neutralize KHP, first we will have to

write the reaction of KHP and NaOH.

NaOH (aq) + KHP (aq) ---> NaKP (aq) + H₂O(l) ;

The mole to mole ratio of NaOH and KHP is 1 :1 .

so, moles of NaOH = 0.00736 mol KHP * 1 mol NaOH/ 1 mol KHP = 0.00736 mol NaOH ;

5) Molarity of NaOH = moles of NaOH / Volume of NaOH in ‘L’

Convert 25.30 mL to ‘L’ and then plug the moles and volume in the formula of molarity.

Volume of NaOH in ‘L’ = 25.30 mL * 1L / 1000mL = 0.02530 L

Molarity of NaOH = 0.00736 mol NaOH / 0.02530 L = 0.291M

Hence the molarity of NaOH is 0.291 M

Part II :

volume of Vinegar, mL 10.00 10.00 10.00

final NaOH volume, mL 25.20 25.35 25.08

initial NaOH volume, mL 0.00 0.00 0.00

vol. NaOH used, mL ________ ________ ________

Calculate Moles of NaOH using the Molarity of NaOH from the previous question

Moles of NaOH used ________ ________ ________

moles of vinegar neutralized ________ ________ ________

1) Volume of NaOH used, mL = Final NaOH volume - Initial NaOH volume

Trial 1 : 25.20mL – 0.00mL = 25.20mL

Trial 2 : 25.35 mL - 0.00 mL = 25.35 mL

Trial 3 : 25.08 mL – 0.00 mL = 25.08 mL

2) Convert volume of NaOH to ‘L’

Trial 1 : 25.20mL * 1L/1000mL = 0.02520 L

Trial 2 : 25.35 mL * 1L / 1000 mL = 0.02535 L

Trial 3: 25.08 mL * 1L / 1000 mL = 0.02508 L

3) To find moles of NaOH

Molarity = moles of NaOH / Volume of NaOH in ‘L’

Trial 1 : 0.291 M = moles of NaOH / 0.02520 L

Arranging the equation for ‘mol of NaOH’ we have

moles of NaOH = 0.291 mol/ L * 0.02520 L = 0.00733 mol

Trial 2 : 0.291 M = moles of NaOH / 0.02535 L

Arranging the equation for ‘mol of NaOH’ we have

moles of NaOH = 0.291 mol/ L * 0.02535 L = 0.00738 mol

Trial 3 : 0.291 M = moles of NaOH / 0.02508 L

Arranging the equation for ‘mol of NaOH’ we have

moles of NaOH = 0.291 mol/ L * 0.02508 L = 0.00730 mol

4) To find the moles of vinegar.

Acetic acid is the componenet in the vinegar which reacts with NaOH.

Reaction of acetic acid and NaOH is given as

NaOH(aq) + CH₃ COOH (aq) ---> CH₃ COO^- Na⁺ (aq) + H₂O(l) ;

the mole ratio of acetic acid and NaOH is 1 mol : 1mol . Hence the moles of vinegar would

also remain same.

Trial 1 : mole of acetic acid = 0.00733 mol NaOH * 1 mol CH₃ COOH / 1 mol NaOH

= 0.00733 mol

Similarly,

Trial 2 : 0.00738 mol and Trial 3 : 0.00730 mol

Please also refer to the images I have uploaded with the answer.

Thank you. :)