In: Chemistry
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
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