A chemist dissolves 716.mg of pure potassium hydroxide in enough water to make up 130.mL cof solution. Calculate the pH of the solution. (The temperature of the solution is 25 degree C.) Be sure your answer has the correct number of significant digits.

13

Explanation:

First, we need to find the number of mole of 716.mg of KOH

Molar Mass of KOH = 39 + 16 +1 = 56g/mol

Mass conc. of KOH = 716.mg

Converting from mg to g,we have

716/1000 = 0.716g

Number of mole = Mass conc. /MM

= 0.716/56 = 0.0128mol

Next, we find the number of mole of KOH in 1L of the solution

0.0128mol of KOH dissolves in 130mL(i.e 0.13L) of solution

Therefore, Xmol of KOH will dissolve in 1L of the solution i.e

Xmol of KOH = (0.0128)/0.13

= 0.098mol/L

Next, we find the [OH^-]

KOH <==> K^+ + OH^-

[KOH] =0.098mol/L

[K^+] = 0.098mol/L

[OH^-] = 0.098mol/L

Next, we find the pOH

pOH= -log [OH^-]

pOH = - log 0.098

pOH = 1

Recall

pH + pOH = 14

pH = 14 — pOH

pH = 14 — 1

pH = 13

Th the pH of the solution is 13.

nacl is added to the aqueous of what? ? in order to get a proper answer you have to specify the exact composition of the aqueous solution.

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