Suppose in the lab you measure the solid NaOH and dissolve it into 100.0 mL of water. You then measure 0.2000 g of KHP (KC8H5O4, 204.22 g/mol) and place it in a clean, dry 100-mL beaker, and then dissolve the KHP in about 25 mL of water and add a couple of drops of phenolphthalein indicator. You titrate this with your NaOH (aq) solution and find that the titration requires 9.53 mL of NaOH (aq).

a. What is the concentration of your NaOH (aq) solution?

b. Determine the number of moles of NaOH (aq) that would be required to titrate 250.00 mL of your Kool-Aid solution.

Question

Konnor Pratt

Chemistry

19 March, 05:16

Suppose in the lab you measure the solid NaOH and dissolve it into 100.0 mL of water. You then measure 0.2000 g of KHP (KC8H5O4, 204.22 g/mol) and place it in a clean, dry 100-mL beaker, and then dissolve the KHP in about 25 mL of water and add a couple of drops of phenolphthalein indicator. You titrate this with your NaOH (aq) solution and find that the titration requires 9.53 mL of NaOH (aq).

a. What is the concentration of your NaOH (aq) solution?

b. Determine the number of moles of NaOH (aq) that would be required to titrate 250.00 mL of your Kool-Aid solution.

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Answers (1)

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Coy · Answer 1

Equation of the reaction:

NaOH (aq) + KHC8H4O4 (aq) - -> KNaC8H4O4 (aq) + H2O (l)

A.

Number of moles = mass/molar mass

= 0.20/204.22

= 9.79 x 10-4 mol KHP

By stoichiometry, 1 mole of NaOH reacts with 1 mole of KHP

= 6.267 x 10-4 mol NaOH

Molar concentration = number of moles/volume

= 9.79 x 10-4/0.00953

= 0.103 M NaOH

B.

Number of moles of kool aid = mass/molar mass

= 3.607/342

= 0.01055 mol

Concentration = 0.01055/0.005

= 2.1 M

C1 * V1 = C2 * V2

Concentration of NaOH = 2.1 * 0.005/0.01079

= 0.977 M

Number of moles of NaOH = 0.0105 mol.