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Nookie1986 [14]
2 years ago
14

Find the hydrogen ion concentration of a saturated solution of calcium hydroxide whose ph is 13. 1

Chemistry
1 answer:
salantis [7]2 years ago
4 0

The pH of the solutions is 12.67.  

     

<h3>What is pH of a Solution?</h3>

pH is the measurement of how acidic or basic a solution is. The range of pH values is from 0 to 14. Lower pH values mean higher acidity, while higher pH means higher basicity. If the pH value is equal to 7, then the solution is neutral.

To calculate the pH of the solution, a few approaches are possible. Here, the pOH of the solution is calculated first using the molar concentration of  OH^{-1} ions. Because pH + pOH = 14, the pH of the solution can then be calculated by subtracting pOH from 14.

Because 2 moles of OH^{-1} ions are formed per mole of Ca(OH)_{2}dissolved, the pOH of a saturated Ca(OH)_{2} solution at  20^{.C} is,  

Concentration (C_{OH}) of OH^{-1} ions,

          C_{OH}   = 0.02335 mol Ca(OH)_{2}/L  X 2 mol OH^{-1} /1 mol Ca(OH)_{2}

            C_{OH} = 0.0467 mol OH^{-1}/L

*pOH of saturated Ca(OH)_{2} solution,

                                   pOH = -log [C_{OH}]

                                   pOH = -log [0.0467 mol OH^{-1}/L ]            

                                  pOH = 1.33  

*pH of saturated Ca(OH)_{2} solution,

                                   pH = 14 - 1.33

                                   pH = 12.67

Hence, the pH of the solutions is 12.67.

Learn more about  pH of the solutions here:

brainly.com/question/6988456

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Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process) to 101.33 kPa, where its temperature is 105°C. W
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Answer:

There are 3 steps of this problem.

Explanation:

Step 1.

Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.

Step 2.

Enthalpy of saturated liquid Haq = 781.124 J/g

Enthalpy of saturated vapour Hvap = 2779.7 J/g

Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g

Step 3.

In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy

So, H1=H2

     H2= (1-x)Haq+XHvap.........1

    Putting the values in 1

    2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}

                        = 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)

1904.976 (J/g) = x1998.576 (J/g)

                     x = 1904.976 (J/g)/1998.576 (J/g)

                     x = 0.953

So, the quality of the wet steam is 0.953

                   

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