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kramer
3 years ago
5

Calcium hydroxide is a strong base but is not very soluble ( Ksp = 5.02 X 10-6 ). What is the pH of a saturated solution of Ca(O

H)2 ?
Chemistry
1 answer:
Misha Larkins [42]3 years ago
7 0

Answer : The pH of a saturated solution is, 12.33

Explanation : Given,

K_{sp} = 5.02\times 10^{-6}

First we have to calculate the solubility of OH^- ion.

The balanced equilibrium reaction will be:

Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-

Let the solubility will be, 's'.

The concentration of Ca^{2+} ion = s

The concentration of OH^- ion = 2s

The expression for solubility constant for this reaction will be,

K_{sp}=[Ca^{2+}][OH^-]^2

Let the solubility will be, 's'

K_{sp}=(s)\times (2s)^2

K_{sp}=(4s)^3

Now put the value of K_[sp} in this expression, we get the solubility.

5.02\times 10^{-6}=(4s)^3

s=1.079\times 10^{-2}M

The concentration of Ca^{2+} ion = s = 1.079\times 10^{-2}M

The concentration of OH^- ion = 2s = 2\times (1.079\times 10^{-2}M)=2.158\times 10^{-2}M

First we have to calculate the pOH.

pOH=-\log [OH^-]

pOH=-\log (2.158\times 10^{-2})

pOH=1.67

Now we have to calculate the pH.

pH+pOH=14\\\\pH=14-pOH\\\\pH=14-1.67=12.33

Therefore, the pH of a saturated solution is, 12.33

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Answer:

Na = 32.4% , % S = 22.6% and %O = 45.0%

Explanation:

% Na = 3.4/10.5. × 100%

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A chemist sets up a chemical reaction but finds that none of the reactant molecules have a required activation energy. What is t
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5 0
3 years ago
Read 2 more answers
A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titratio
Anna11 [10]

Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol

nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

                    NaOH       +       HCl       ⇒       NaCl      +         H₂O

Initial          2.5 × 10⁻³         1.5 × 10⁻³               0                      0

Reaction    -1.5 × 10⁻³        -1.5 × 10⁻³          1.5 × 10⁻³          1.5 × 10⁻³

Final            1.0 × 10⁻³               0                 1.5 × 10⁻³          1.5 × 10⁻³

The concentration of NaOH is:

[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

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3 years ago
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babunello [35]

Answer:

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Explanation:

A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.

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Cell voltage= -1.19 V - (-0.14V)

Cell voltage= -1.05 V

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