Question 13.3: Consider barium hydroxide, Ba(OH)2, a white powdery substanc...
Consider barium hydroxide, Ba(OH)_{2}, a white powdery substance used in the treatment of waste water. A student is asked by his TA to prepare 455 mL of a solution containing 4.23 g of Ba(OH)_{2}.
a What is the calculated pH of the prepared solution?
ANALYSIS
Information given: mass Ba(OH)_{2} (4.23 g); volume of solution (455 mL)
Information implied: molar mass of Ba(OH)_{2}; K_{w}
Asked for: pH of the solution
STRATEGY
1. Start by expressing the concentration in g Ba(OH)_{2}/L of solution.
2. Follow the pathway:
mass\> Ba(OH)_{2}/L \xrightarrow{MM} [Ba(OH)_{2}]\xrightarrow{2[OH^-]/[Ba(OH)_{2}]} [OH^-] \xrightarrow{K_{w}} [H^{+}] \xrightarrow{Eq \>13.3} pH
pH = -\log_{10}[H^+] = -\log_{10}[H_{3}O^{+}] (13.3)
or
[H^+] = [H_{3}O^+] = 10^{-pH}b His TA measures the pH of the prepared solution with a pH meter and finds it to be 12.91. How much Ba(OH)_{2} did the student really add?
ANALYSIS
Information given: pH (12.91) mass Ba(OH)_{2} to be added (4.23 g); volume of solution (455 mL)
Information implied: molar mass of Ba(OH)_{2}; K_{w}
Asked for: mass Ba(OH)_{2} added compared to student’s calculation
STRATEGY
The pathway to follow is the reverse of that in (a):
pH \xrightarrow{Eq\> 13.3} [H^{+}] \xrightarrow{K_{w}} [OH^{-}] \xrightarrow{2[OH^{-}]/[Ba(OH)_{2}]} [Ba(OH)_{2}] \xrightarrow{V} mol\> Ba(OH)_{2} \xrightarrow{MM} mass
c The TA added 0.50 g of NaOH to the prepared solution to increase the pH. What is the pH of the prepared solution after the addition of NaOH?
ANALYSIS
Information given: from (b) [OH^{-}] due to Ba(OH)_{2} (0.081 M); volume of solution (455 mL) mass of NaOH added (0.50 g)
Information implied: molar mass of NaOH; K_{w}
Asked for: pH of the solution
STRATEGY
1. Find moles OH^{-} contributed by Ba(OH)_{2} from (b).
2. Find moles OH^{-} contributed by NaOH.
3. Find [OH^{-}] after NaOH addition.
\frac{ (mol\> OH^{-})_{NaOH} + (mol\> OH^{-})_{Ba(OH)_{2}}}{V}
4. Find [H^{+}] and pH.
