How concentrated can lye solution be before its ph rises above calcium hydroxide saturated water?
1 answer:
The solubility of calcium hydroxide (Ca(OH)2) in water at 
is 0.02335 <span>mol/L. When a Ca(OH)2 solution has a concentration of 0.02335 mol/L at the said temperature, a saturated solution is formed.
Upon dissolution and dissociation of 1 mole Ca(OH)2, 1 mole of calcium ion (</span>
) and 2 moles of hydroxide ions (
) are formed. The concentration of the hydroxide ions determine the pH of the solution. This same concentration would be needed in calculating the concentration of lye solution needed.
1. 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 ions. Because pH + pOH = 14, the pH of the solution can then be calculated by subtracting pOH from 14.
Because 2 moles of ions are formed per mole of Ca(OH)2 dissolved, the pOH of a saturated Ca(OH)2 solution at is,
*Concentration (
) of ions,
= 0.02335 mol Ca(OH)2/L * 2 mol / 1 mol Ca(OH)2
= 0.0467 mol /L
*pOH of saturated Ca(OH)2 solution,
pOH = -log []
pOH = -log [0.0467 mol/L ]
pOH = 1.33
*pH of saturated Ca(OH)2 solution,
pH = 14 - 1.33
pH = 12.67
2. To find the concentration of lye (NaOH) solution that would give the same pH as a saturated Ca(OH)2 solution at the same temperature, the pOH of the saturated Ca(OH)2 solution will be used.
For every mole of NaOH that dissolves and dissociates, 1 mole of ions is formed. Thus,
*Concentration (
) of ions,
pOH = -log []
1.33 = -log [ ]
= 10^-1.33
= 0.0467 mol / L
Because 1 mole of NaOH produces only 1 mole of ions, the concentration of the ions should then be equal to the concentration of the lye (NaOH) solution. Thus, 0.0467 mol/L of lye would give the same pH as that of 0.0233 mol/L (saturated) Ca(OH)2 solution at . The pH of both solutions is 12.67.
Upon dissolution and dissociation of 1 mole Ca(OH)2, 1 mole of calcium ion (</span>
1. 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
Because 2 moles of
*Concentration (
*pOH of saturated Ca(OH)2 solution,
pOH = -log [
pOH = -log [0.0467 mol
pOH = 1.33
*pH of saturated Ca(OH)2 solution,
pH = 14 - 1.33
pH = 12.67
2. To find the concentration of lye (NaOH) solution that would give the same pH as a saturated Ca(OH)2 solution at the same temperature, the pOH of the saturated Ca(OH)2 solution will be used.
For every mole of NaOH that dissolves and dissociates, 1 mole of
*Concentration (
pOH = -log [
1.33 = -log [
Because 1 mole of NaOH produces only 1 mole of
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Answer:
Value of equilibrium constant is 0.0888
Explanation:
Both and
are gaseous. Hence equilibrium constant depends upon partial pressures of
and
.
Initially no and
were present.
Hence mole fraction of and
at equilibrium can be calculated from coefficient of
and
in balanced equation.
Mole fraction of = (number of moles of
)/(total number of moles of
and
) =
Mole fraction of = (number of moles of
)/(total number of moles of
and
) =
Let's assume both and
behaves ideally.
Therefore partial pressure of ,
and P_{CO_{2}}= x_{CO_{2}}.P_{total}
Where x represents mole fraction
So,
So,