Calculate the molarity (M) of a solution of the strong base Ca(OH)2 if the pOH is 0.650.
pOH= 0.650
pH = 14-0.650
pJ=13.35
-log(13.35) = -1.13
pOH = 0.650
-0.650 = log(OH^-)
(OH^-) = 4.47 M
Ca(OH)2 = 1/2 that or 2.23 M
To calculate the molarity (M) of a solution of Ca(OH)2, we need to first understand the relationship between pOH and pH.
pOH is a measure of the basicity of a solution and is defined as the negative logarithm (base 10) of the hydroxide ion concentration [OH-]. It is related to pH through the equation: pOH + pH = 14.
In this case, we are given that the pOH is 0.650. To find the pH, we subtract the pOH from 14:
pH = 14 - 0.650
pH ≈ 13.35
Now that we have the pH, we can proceed to find the hydroxide ion concentration [OH-] using the equation:
[OH-] = 10^(-pOH)
Substituting the given pOH value:
[OH-] = 10^(-0.650)
[OH-] ≈ 0.244 M
Next, we need to determine the concentration of the calcium hydroxide, Ca(OH)2. Since calcium hydroxide dissociates into two hydroxide ions, the concentration of Ca(OH)2 will be double the concentration of hydroxide ions.
[Ca(OH)2] ≈ 2 * [OH-]
[Ca(OH)2] ≈ 2 * 0.244 M
[Ca(OH)2] ≈ 0.488 M
Therefore, the molarity (M) of the solution of Ca(OH)2 is approximately 0.488 M.