Question

In: Chemistry

In 20.0 mL deionized H2O + 5.0 mL NaOH (concentration 0.1105 M) a) How many moles...

In 20.0 mL deionized H2O + 5.0 mL NaOH (concentration 0.1105 M)

a) How many moles of NaOH are in Solution B after it is mixed?

b)What is the concentration of NaOH in Solution B after it is mixed?

c)Show a complete calculation of how you calculated the pH of Solution B after mixing.

d)Calculated pH of solution B

Solutions

Expert Solution

Ans. #a. Moles of NaOH = Molarity x Volume of solution in liters

                                                = 0.1105 M x 0.005 L

                                                = 0.0005525 mol

Hence, number of moles of NaOH = 0.0005525 mol in final solution.

Note: Dilution does not change the number of moles of NaOH (though it changes concertation).

#b. Final volume of mixed solution = 20.0 mL (water) + 5.0 mL (NaOH soln.)

                                                            = 25.0 mL

                                                            = 0.025 L

Now,

            [NaOH] = Moles of NaOH / Volume of mixed solution in liters

                                    = 0.0005525 mol / 0.025 L

                                    = 0.0221 M

#c. NaOH is a strong base. 1 mol NaOH completely dissociate to yield 1 mol OH- ions as follow-

            NaOH(aq) + H2O(l) --------> Na+ + H2O(aq) + OH-(aq)

So,

            [OH-] in mixed solution = 0.0221 M = [NaOH]

# Using           [H3O+] [OH-] = 10-14

            Or, [H3O+] = 10-14 / [OH-] = 10-14 / 0.0221 = 4.5249 x 10-13 M

# Now,

            pH of mixed solution = -log [H3O+] = -log (4.5249 x 10-13) = 12.34

#d. For solution B; 0.1105 M NaOH

Using [H3O+] [OH-] = 10-14

            Or, [H3O+] = 10-14 / 0.1105 = 9.0498 x 10-14 M

# Now,

            pH of solution B = -log (9.0498 x 10-14) = 13.04


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