Calculate the pH change that results when 11 mL of 5.7 M NaOH is added to 783 mL
ID: 956939 • Letter: C
Question
Calculate the pH change that results when 11 mL of 5.7 M NaOH is added to 783 mL of each the following solutions. Use the Acid-Base Table.
(a) pure water WebAssign will check your answer for the correct number of significant figures.
(b) 0.10 M NH4Cl WebAssign will check your answer for the correct number of significant figures.
(c) 0.10 M NH3 WebAssign will check your answer for the correct number of significant figures.
(d) a solution that is 0.10 M in each NH4+ and NH3 WebAssign will check your answer for the correct number of significant figures.
Explanation / Answer
(a) pure water
NaOH concentration = 11 x 5.7 / (11 + 783) = 0.079M
pOH = -log [OH-] = -log (0.079) = 1.10
pH + pOH = 14
pH = 12.9
(b) 0.10 M NH4Cl
millimoles of NaOH = 11 x 5.7 = 62.7
millimoles of NH4Cl = 0.10 x 783 = 78.3
NH4+ + OH- --------------------> NH3 + H2O
78.3 62.7 0 0
15.6 0 62.7
pOH = pKb + log [NH4+ / NH3]
pOH = 4.74 + log (15.6 / 62.7)
pOH = 4.14
pH = 9.86
(c) 0.10 M NH3
[OH-] from NH3 = sqrt (Kb x C) = sqrt (1.8 x 10^-5 x 0.1) = 1.34 x 10^-3 M
[OH-] from NaOH = 5.7 M
[OH-] total = (5.7 x 11 + 1.34 x 10^-3 x 783 ) / ( 783 + 11 )
= 0.080 M
pOH = 1.10
pH = 12.9
(d) a solution that is 0.10 M in each NH4+ and NH3
C = 11 x 5.7 = 62.7
pOH = pKb + log [NH4+ - C / NH3 + C]
pOH = 4.74 + log (78.3 - 62.7 / 78.3 + 62.7)
pOH = 3.78
pH = 10.2