What is the pH of a solution prepared by mixing 47 mL of 0.073 M Ba(OH)2 and 54 mL of 0.20 M KOH? (This is just a mixing problem: simply add up the mmol OH- and divide by the total mL.)

Answer : The pH of the solution is, 13.2

Explanation :

First we have to calculate the mole of and .

and,

Now we have to calculate the moles of hydroxide ion.

As, 1 mole of dissociates to give 1 mole of barium ion and 2 mole of hydroxide ion.

So, 0.00343 mole of dissociates to give 0.00343 mole of barium ion and (2×0.00343=0.00686) mole of hydroxide ion.

and,

As, 1 mole of dissociates to give 1 mole of potassium ion and 1 mole of hydroxide ion.

So, 0.0108 mole of dissociates to give 0.0108 mole of potassium ion and 0.0108 mole of hydroxide ion.

Now we have to calculate the total moles of hydroxide ion and volume of solution.

Total moles of chloride ion = 0.00686 + 0.0108 = 0.01766 mol

Total volume of solution = 47 mL + 54 mL = 101 mL = 0.101 L

Now we have to calculate the molarity of chloride ion in the final solution.

Now we have to calculate the value of pOH of the solution.

Now we have to calculate the value of pH of the solution.

Therefore, the pH of the solution is, 13.2

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.