Calculate the Ksp for Ca(OH)2 for each trial Table 1. Room Temperature Raw DataTrial 1Trial 2Temperatue ("C)21.5 °C21.5 °CConcentration of HCl solution (M) 0.01530.0153Sample aliquot volume (mL)10.0010.00Initial buret reading (mL)0.0016.20Final buret reading at equivalencepoint (mL)16.2033.35Table 2. ~100°C Raw DataTrial 3Trial 4Temperature (°C)94.0°C94.0°CConcentration of HCl solution (M) 0.01530.0153Sample aliquot volume (mL)|10.0010.00Initial buret reading (mL)0.0013.80Final buret reading at equivalencepoint (mL)13.8026.05

The dissociation of calcium hydroxide is as follows. CaOH2s⇔Ca2+aq+2OH-aq That means concentration…

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

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Calculate the Ksp for Ca(OH)2 for each trial

Table 1. Room Temperature Raw Data
Trial 1
Trial 2
Temperatue ("C)
21.5 °C
21.5 °C
Concentration of HCl solution (M) 0.0153
0.0153
Sample aliquot volume (mL)
10.00
10.00
Initial buret reading (mL)
0.00
16.20
Final buret reading at equivalence
point (mL)
16.20
33.35
Table 2. ~100°C Raw Data
Trial 3
Trial 4
Temperature (°C)
94.0°C
94.0°C
Concentration of HCl solution (M) 0.0153
0.0153
Sample aliquot volume (mL)
|10.00
10.00
Initial buret reading (mL)
0.00
13.80
Final buret reading at equivalence
point (mL)
13.80
26.05

Expert Solution

Step 1

The dissociation of calcium hydroxide is as follows.

$Ca {(OH)}_{2 (s)} \Leftrightarrow {Ca}^{2 +}_{(aq)} + 2 {OH}^{-}_{(aq)}$

That means concentration of calcium ion and hydroxide ion are in $1 : 2$ ratio.

$\begin{array}{rcl} [{Ca}^{2 +}_{(aq)}] \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} 2 \end{array} \begin{array}{rcl} [{OH}^{-}_{(aq)}] \end{array}$

The $K_{sp}$ expression for this reaction is as follows.

$\begin{array}{rcl} K_{sp} & = & [{Ca}^{2 +}_{(aq)}] {[{OH}^{-}_{(aq)}]}^{2} \\ = & s \times {(2 s)}^{2} \\ = & 4 s^{3} \end{array}$

Step 2

The equation for the reaction between $HCl$ and $Ca {(OH)}_{2}$ is as follows.

$Ca {(OH)}_{2} + 2 HCl \to 2 H_{2} O + {CaCl}_{2}$

The no of moles $H^{+} (HCl)$ reacted in the trial 1 can be calculated as follows.

$\begin{array}{rcl} Concentration of HCl & = & 0.0153 M \\ = & 0.0153 mol / L \\ Volume of HCl used & = & final buret reading - initial buret reading \\ = & 16.20 mL \\ no of moles of HCl used & = & concentration \times volume (L) \\ = & 0.0153 mol / L \times (16.20 \times 10^{- 3} L) \\ = & 2.4786 \times 10^{- 4} moles \end{array}$

The no of moles of ${OH}^{-}$ can be calculated as follows.

$\begin{array}{rcl} 2 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} 4786 \end{array} \begin{array}{rcl} \times \end{array} \begin{array}{rcl} 10 \end{array} \begin{array}{rcl} ^{- 4} mole HCl \times \frac{1 mol Ca {(OH)}_{2}}{2 mol HCl} \times \frac{2 mol {OH}^{-}}{1 mol Ca {(OH)}_{2}} & = \end{array} \begin{array}{rcl} 2 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} 4786 \end{array} \begin{array}{rcl} \times \end{array} \begin{array}{rcl} 10 \end{array} \begin{array}{rcl} ^{- 4} \end{array} \begin{array}{rcl} mol \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} Ca \end{array} \begin{array}{rcl} (OH) \end{array} \begin{array}{rcl} _{2} \end{array} \begin{array}{rcl} \end{array}$

That means we have $\begin{array}{rcl} 2 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} 4786 \end{array} \begin{array}{rcl} \times \end{array} \begin{array}{rcl} 10 \end{array} \begin{array}{rcl} ^{- 4} \end{array} \begin{array}{rcl} mol \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} OH \end{array} \begin{array}{rcl} \end{array}$ ^- in 10 mL.

The concentration/molarity is defined for $1 L$ .

$\begin{array}{rcl} 2.4786 \times 10^{- 4} mol {OH}^{-} in 10 mL \\ \frac{2.4786 \times 10^{- 4}}{10 mL} mol {OH}^{-} in 1 mL \\ (\frac{2.4786 \times 10^{- 4}}{10 mL} \times 1000) mol {OH}^{-} in 1000 mL \\ 0.024786 mol {OH}^{-} in 1000 mL \\ 0.024786 mol {OH}^{-} in 1 L \end{array}$

Now we can calculate $K_{sp}$ for trial 1.

$\begin{array}{rcl} K_{sp} & = & 4 s^{3} \\ = & 4 {(0.024786)}^{3} \\ = & 6.09 \times 10^{- 5} \end{array}$

Hence the $K_{sp}$ for trial 1 is $6.09 \times 10^{- 5}$

Step 3

The equation for the reaction between $HCl$ and $Ca {(OH)}_{2}$ is as follows.

$Ca {(OH)}_{2} + 2 HCl \to 2 H_{2} O + {CaCl}_{2}$

The no of moles $H^{+} (HCl)$ reacted in the trial 2 can be calculated as follows.

$\begin{array}{rcl} Concentration of HCl & = & 0.0153 M \\ = & 0.0153 mol / L \\ Volume of HCl used & = & final buret reading - initial buret reading \\ = & (33.35 - 16.20) mL \\ = & 17.15 mL \\ no of moles of HCl used & = & concentration \times volume (L) \\ = & 0.0153 mol / L \times (17.15 \times 10^{- 3} L) \\ = & 2.62395 \times 10^{- 4} moles \end{array}$

The no of moles of ${OH}^{-}$ can be calculated as follows.

$\begin{array}{rcl} 2 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} 62395 \end{array} \begin{array}{rcl} \times \end{array} \begin{array}{rcl} 10 \end{array} \begin{array}{rcl} ^{- 4} mole HCl \times \frac{1 mol Ca {(OH)}_{2}}{2 mol HCl} \times \frac{2 mol {OH}^{-}}{1 mol Ca {(OH)}_{2}} & = \end{array} \begin{array}{rcl} 2 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} 62395 \end{array} \begin{array}{rcl} \times \end{array} \begin{array}{rcl} 10 \end{array} \begin{array}{rcl} ^{- 4} \end{array} \begin{array}{rcl} mol \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} Ca \end{array} \begin{array}{rcl} (OH) \end{array} \begin{array}{rcl} _{2} \end{array} \begin{array}{rcl} \end{array}$

That means we have $2.62395 \times 10^{- 4} mol {OH}^{-} in 10 mL .$

The concentration/molarity is defined for $1 L$ .

$\begin{array}{rcl} 2.62395 \times 10^{- 4} mol {OH}^{-} in 10 mL \\ \frac{2.62395 \times 10^{- 4}}{10 mL} mol {OH}^{-} in 1 mL \\ (\frac{2.62395 \times 10^{- 4}}{10 mL} \times 1000) mol {OH}^{-} in 1000 mL \\ 0.0262395 mol {OH}^{-} in 1000 mL \\ 0.0262395 mol {OH}^{-} in 1 L \end{array}$

Now we can calculate $K_{sp}$ for trial 2.

$\begin{array}{rcl} K_{sp} & = & 4 s^{3} \\ = & 4 {(0.0262395)}^{3} \\ = & 7.23 \times 10^{- 5} \end{array}$

Hence the $K_{sp}$ for trial 2 is $7.23 \times 10^{- 5}$

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