Calculate the molar solubility of 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 balanced reaction of Ca(OH)2 with HCl is as follows: Ca(OH)2(aq)+2HCl(aq)→CaCl2(aq)+2H2O(l)…

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Chemistry

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ISBN:9781305957404

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

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

Chapter1: Chemical Foundations

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Calculate the molar solubility of 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 balanced reaction of Ca(OH)₂ with HCl is as follows:

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

Volume of Ca(OH)₂ taken = 10.0 mL

According to the balanced reaction, 1 mol of Ca(OH)₂ reacts with 2 mol of HCl

Step 2

Trial 1:

Calculation of no. of mol of HCl:

$\begin{array}{rcl} n & = & Molarity \times Volume \\ = & 0.0153 mol / L \times (16.20 - 0.00) \times 10^{- 3} L \\ = & 0.24786 \times 10^{- 3} mol \end{array}$

Calculation of no. of mol of Ca(OH))₂:

$\begin{array}{rcl} n & = & \frac{1}{2} \times 0.24786 \times 10^{- 3} mol \\ = & 0.12393 \times 10^{- 3} mol \end{array}$

Calculation of molarity of Ca(OH)₂:

$\begin{array}{rcl} Molarity & = & \frac{No . of mol}{Volume} \\ = & \frac{0.12393 \times 10^{- 3} mol}{10 \times 10^{- 3} L} \\ = & 0.012393 mol / L \end{array}$

Calculation of solubility (g/L):

$\begin{array}{rcl} Solubility & = & 0.012393 mol / L \times 74.093 g / mol \\ = & 0.9182 g / L \end{array}$

Thus solubility of Ca(OH)₂ is 0.9812 g/L

Step 3

Trial 2:

Calculation of no. of mol of HCl:

$\begin{array}{rcl} n & = & Molarity \times Volume \\ = & 0.0153 mol / L \times (33.35 - 16.20) \times 10^{- 3} L \\ = & 0.2624 \times 10^{- 3} mol \end{array}$

Calculation of no. of mol of Ca(OH))₂:

$\begin{array}{rcl} n & = & \frac{1}{2} \times 0.2624 \times 10^{- 3} mol \\ = & 0.1312 \times 10^{- 3} mol \end{array}$

Calculation of molarity of Ca(OH)₂:

$\begin{array}{rcl} Molarity & = & \frac{No . of mol}{Volume} \\ = & \frac{0.1312 \times 10^{- 3} mol}{10 \times 10^{- 3} L} \\ = & 0.01312 mol / L \end{array}$

Calculation of solubility (g/L):

$\begin{array}{rcl} Solubility & = & 0.01312 mol / L \times 74.093 g / mol \\ = & 0.9721 g / L \end{array}$

Thus solubility of Ca(OH)₂ is 0.9721 g/L

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