Question

At 25 °C, how many dissociated OH– ions are there in 1243 mL of an aqueous solution whose pH is 1.60?

Answer 1

Expression for pOH can be obtained as follows:

pH + pOH   = 14

pOH = 14 - pH

pOH = 14 – 1.60

pOH = 12.4

Concentration of OH- ions can be calculated using following expression:

pOH = -log[OH-]

12.4 = -log[OH-]

[OH-] = 0.3981 x 10^-12 mol/L

Therefore, number of moles of OH- ions present in 1243mL of given solution can be calculated as follows:

Number of moles of OH- = Molarity x volume

Number of moles of OH- = 0.3981 x 10^-12 mol/L x 1.243L

Number of moles of OH- = 0.4948 x 10^-12 mol

Since, one mole of ions contains 6.022 x 10²³ number of ions (Avagadro’s Number), the number of ions in 0.4948 x 10^-12 moles of ions can be calculated as follows:

Number of ions = 6.022 x 10²³ x 0.4948 x 10^-12

Number of ions = 2.980 x 10¹¹

Therefore, number of OH- ions in given aqueous solution is 2.980 x 10¹¹