Need help figuring this out. For each trial, calculate the mass of your unknown

Question

Need help figuring this out.

For each trial, calculate the mass of your unknown acid required to neutralize one mole of hydroxide ion.

Using the calculated mass of acid/mole OH- values, determine the percent relative average deviation.

Report the average grams acid/mole hydroxide neutralized for your unknown

Trial NumberKHP used (g) Initial Buret Reading (mL) Final Buret Reading (mL) Mass of Color of Volume of NaOH Ratlo of Mass to Volume (g/ml 157.53 160.37 159.06 Endpoint used (mL) Pink 18.32 0.1149 0.1653 0.1214 0.22 26.74 Light Pink 19.63 Light Pink 18.1 26.51 19.31 0.32

Explanation / Answer

The balanced chemical equation for the reaction is given as

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

As per the stoichiometric equation,

1 mole KHP = 1 mole NaOH

Molar mass of KHP = 204.22 g/mol

Trial Number

Mass of KHP used (g)

Mole(s) of KHP used = (mass of KHP)/(molar mass of KHP)

Mole(s) of NaOH neutralized = moles of KHP taken (as per the stoichiometric equation)

Mass of KHP required to neutralize 1 mole of NaOH (g) = (mass of KHP)*(1 mole KHP)/(moles of KHP)

1

0.1149

(0.1149)/(204.22) = 5.6263*10-4

5.6263*10-4

203.857

2

0.1653

(0.1653)/(204.22) = 8.0942*10-4

8.0942*10-4

204.220

3

0.1214

(0.1214)/(204.22) = 5.9446*10-4

5.9446*10-4

204.219

Trial Number

Initial Volume of NaOH (mL)

Final Volume of NaOH (mL)

Volume of NaOH (mL)

Mass of KHP (g) to neutralize 1 mole NaOH

Mass to Volume (g/mL)

1

0.22

18.32

18.10

203.857

11.263

2

0.23

26.74

26.51

204.220

7.703

3

0.32

19.63

19.31

204.219

10.57

Please clarify the remaining questions; its difficult to understand. Please provide a snapshot of the question. I believe you want to find out the average and standard deviation of the masses of KHP required to neutralize 1 mole of NaOH.

Average mass of KHP = 1/3*(203.857 + 204.220 + 204.219) g = 204.099 g (ans).

Standard deviation = [(203.857 – 204.099)2 + (204.220 – 204.099)2 + (204.219 – 204.099)2]/(3 – 1) = 0.209 g (ans).

Trial Number

Mass of KHP used (g)

Mole(s) of KHP used = (mass of KHP)/(molar mass of KHP)

Mole(s) of NaOH neutralized = moles of KHP taken (as per the stoichiometric equation)

Mass of KHP required to neutralize 1 mole of NaOH (g) = (mass of KHP)*(1 mole KHP)/(moles of KHP)

1

0.1149

(0.1149)/(204.22) = 5.6263*10-4

5.6263*10-4

203.857

2

0.1653

(0.1653)/(204.22) = 8.0942*10-4

8.0942*10-4

204.220

3

0.1214

(0.1214)/(204.22) = 5.9446*10-4

5.9446*10-4

204.219

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