A. A titration of a 0.7871g sampe KHP (204.23g/mol) required 21.22ml of NaOh sol

Question

A. A titration of a 0.7871g sampe KHP (204.23g/mol) required 21.22ml of NaOh solution to reach the endpoint. What is the molarity of the sodium hydroxide solution?

B. Three more samples of KHP weighing 0.8001g, 0.7001g, and 0.7666g were titrated with the same NaOH solution. The volumes for the titrations were 21.55ml, 17.44ml, and 18.37 ml respectively. Calculate the molarity and then calculate the mean for all measurements.

C. Test precsison for these measurements. Can you reject any results using the Q test?

D. Calculate the standard deviation, 95 confidence limit for the four titrations if the Q test showed that you cant reject a data value or calculate the mean, standard deviation, and 95% confidence limit for the best three trials if rejection was allowed.

Explanation / Answer

The calculations are as follows,

We will use the following relations for calculation

moles = g/molar mass

molarity = moles/L

Standard data

Molar mass of NaOH = 40 g/mol

Molar mass of KHP = 204.23 g/mol

A. first calculate moles of KHP,

Given data

g of KHP = 0.7871 g

mL of NaOH used for titration = 21.22 mL = 0.02122 L

So, moles of KHP = 0.7871/204.23

                              = 3.854 x 10^-3 M

Since one mole of KHP is reacts with 1 mole of KHP.

thus moles of NaOH = 3.854 x 10^-3 M

Molarity of NaOH solution = 3.854 x 10^-3 / 0.02122

                                          = 0.18 M

B. KHP weighing (a) 0.8001g, (b) 0.7001g, and (c) 0.7666g

(a) 0.8001 g of KHP

moles of KHP = 0.8001 / 204.23

                       = 3.92 x 10^-3 M

This will be the moles of NaOH = 3.92 x 10^-3 M

Volume of NaOH used for titration = 21.55 ml

                                                       = 0.02155 L

Molarity of NaOH solution = 3.92 x 10^-3 x 0.02155

                                          = 0.18 M

(b) 0.7001 g of KHP

moles of KHP = 0.7001 / 204.23

                       = 3.43 x 10^-3 M

This will be the moles of NaOH = 3.43 x 10^-3 M

Volume of NaOH used for titration = 17.44 ml

                                                       = 0.01744 L

Molarity of NaOH solution = 3.43 x 10^-3 x 0.01744

                                          = 0.197 M

(c) 0.7666 g of KHP

moles of KHP = 0.7001 / 204.23

                       = 3.75 x 10^-3 M

This will be the moles of NaOH = 3.43 x 10^-3 M

Volume of NaOH used for titration = 18.37 ml

                                                       = 0.01837 L

Molarity of NaOH solution = 3.43 x 10^-3 x 0.01837

                                          = 0.204 M

Mean molarity of NaOH solution = 0.18 + 0.197 + 0.204 / 3

                                                    = 0.194 M

C. All the results are within the error limits.

D. standard deviation If 0.204 M is rejectable data will be,

    Find deviation for each point,

(0.18 - 0.194)^2 = 1.96 x 10^-4

(0.18 - 0.194)^2 = 1.96 x 10^-4

(0.197 - 0.194)^2 = 9 x 10^-6

Find variance,

1.96 x 10^-4 + 1.96 x 10^-4 + 9 x 10^-6 / 3

= 1.34 x 10^-4

standard deviation = sq.rt.(variance)

                              = sq.rt.(1.34 x 10^-4)

                              = 0.012

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