Chemistry - pH Question?

what is the pH of 500mL of soln. containing 0.0124 gram of Ca(OH)2.

If anyone can help me set this up that would be great!

Convert to molarity.

0.0124 g / molecular mass of Ca(OH)2 / 0.5 liter

Assume all is ionized. You multiply the molarity by 2, because each mole of Ca(OH)2 gives 2 moles of OH-.

Now you have the moles of OH-. Calculate the pOH in the usual way. Subtract from 14 and you have pH.
First, convert Ca(OH)2 to moles:

Ca(OH)2 is 40.1 + 2(16) + 2(1.01) = 74.1 g/mol

.0124 g x 1 mol/74.1g = 1.67 x 10^-4 mol

Second, we have to divide that in half to find the molarity because you have half of a liter. (500mL)

1.67 x 10^-4 / 2 = 8.35 x 10^-5 M

Third, we know that there are two OH- for every Ca(OH)2 assuming that it completely ionizes, so we multiply that by two, which obviously brings us back to 1.67 x 10^-4 for our concentration of OH-.

Fourth, we can find the pOH by finding the negative log of 1.67 x 10^-4, which = 3.77

Fifth, we can subtract 3.77 from 14 to find the pH, which finally equals 10.23, which is appropriate for a strong base.
The first thing is to find how many moles of Ca(OH)2 you have, from which you can find the molarity of Ca(OH)2 concentration (moles/liter) you have. Ca(OH)2 is a strong base so all of the OH- is floating around as free ions and none is bound up in molecular form. Because there are 2 moles of OH in every mole of Ca(OH)2, you double this to find the concentration of OH-, also known as [OH-]. The product [H+] * [OH-] is equal to 10^(-14). From this, you can calculate [H+] = 10^(-14) / [OH-].
pH is then -log [H+]

Once you have [OH-], you can also figure pH by first calculating pOH, which is -log [OH-]. Because pH + pOH = 14, pH = 14 - pOH.

[post hoc] ksquared013's molar mass and number of moles calculation is correct, but his calculation for concentration of Ca(OH)2 is not. You must divide the molar quantity of solute by the volume of solution, not multiply.

The answers post by the user, for information only, does not guarantee the right.

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