0.4857 g of KHP was dissolved in water and titrated with a solution NaOH of unknown...

0.4857 g of KHP was dissolved in water and titrated with a solution NaOH of unknown Molarity. A volume of 20.35 mL of NaOH solution was required to reach the endpoint. Two more runs with 0.4449 g and 0.4701 g of KHP required 19.04 and 19.73 mL of the NaOH respectively.

a)Calculate the mL of NaOH/g KHP for each run.

b)Find the mean, standard deviation, and CV

c) The fourth run with 0.4501 g of KHP needed 18.89 mL of NaOH, find the new mean, standard deviation, and CV

d)What is the molarity of the NaOH solution?

Expert Solution

In the question, it is not mentioned which mean we must take so we have computed the expected value in that case.

Please refer to the solution below.

queen_honey_blossom answered 2 years ago

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