In this lab I took 25mL of solid and saturated Ca(OH) 2 and added HCl solution u

Question

In this lab I took 25mL of solid and saturated Ca(OH)2 and added HCl solution until the pH was measured below 5.

Solid

First titration was 21.5 celsius

second was 22.5 celsius

Saturated

third was 52.5 celsius

and fourth was 37.0 celsius

What is the question asking?

here were the results in order of above

Run 1 volume pH v p ml 1 12.87 2 12.83 3 12.84 4 12.79 8 12.66 12 12.45 15 12.31 18 11.94 21 11.47 23 11.21 24 11.08 26 9.66 28 7.78 28.5 4.69 Run 2 volum pH v p ml 0 12.65 7 12.25 10 11.85 13 9.75 14.8 6.73 15.5 4.52 Run 3 volum pH v p ml 0 12.6 3 12.53 7 12.23 10 11.7 12 10.92 13 10.19 14 8.26 15 4.97 Run 4 volum pH v p ml 0 12.77 4 12.68 8 12.54 12.3 12.38 15 12.2 18 11.82 21 10.81 23 7.55 24 4.89

Explanation / Answer

Solution:

Let, s be the solubility of Ca(OH)2.

Ca(OH)2 <------> Ca2+ + 2OH-

     s                       s         2s

Solubility product can be given by the following equation:

Ksp = [Ca2+][OH-]2

      = sx(2s)2 = 4s3

For run-1: To calculate the initial concentration of [OH-] at 22.5oC, pH at v=0 mL is required. This is missing in the titration table for run-1.

For run-2: Initial concentration (at v =0 mL) of [OH-] at 22.5oC can be calculated as follows:

pOH = -log[OH-]

14 – pH = -log[OH-]

14 – 12.65 = -log[OH-]

-1.35 = log[OH-]

[OH-]= s =10-1.35 = 0.0446 moles/L

Therefore molar solubility of Ca(OH)2 = 2[OH-] moles/L = 2x0.0446 =0.09 moles/L (since, Ca(OH)2 contains 2 moles of OH-)

Ksp = [Ca2+][OH-]2 = sx(2s)2 = 4s3 = 4x(0.0446)3 = 3.5x10-4 mol3L-3

For run-3: Initial concentration (at v =0 mL) of [OH-] at 52.5oC can be calculated as follows:

pOH = -log[OH-]

14 – pH = -log[OH-]

14 – 12.6 = -log[OH-]

-1.4 = log[OH-]

[OH-]= s =10-1.4 = 0.0398 moles/L

Therefore molar solubility of Ca(OH)2 = 2[OH-] moles/L = 2x0.0398 =0.08 moles/L (since, Ca(OH)2 contains 2 moles of OH-)

Ksp = [Ca2+][OH-]2 = sx(2s)2 = 4s3 = 4x(0.0398)3 = 2.5x10-4 mol3L-3

For run-4: Initial concentration (at v =0 mL) of [OH-] at 37oC can be calculated as follows:

pOH = -log[OH-]

14 – pH = -log[OH-]

14 – 12.77 = -log[OH-]

-1.23 = log[OH-]

[OH-]= s =10-1.23 = 0.0588 moles/L

Therefore molar solubility of Ca(OH)2 = 2[OH-] moles/L = 2x0.0588 =0.11 moles/L (since, Ca(OH)2 contains 2 moles of OH-)

Ksp = [Ca2+][OH-]2 = sx(2s)2 = 4s3 = 4x(0.0588)3 = 8.1x10-4 mol3L-3

Now, average, s = total of ‘s’ calculated in each temperature condition/no. of times calculated

                           = (0.0446+0.0398+0.0588)/3 = 0.0477 moles/L

Similarly, average Ksp = (3.5+2.5+8.1)x10-4/3 = 4.7x10-4 mol3L-3

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