Question

Determine the pOH (to two decimal places) of the solution that is produced by mixing 5.83 mL of 4.19×10-2 M Mg(OH)2 with 918 mL of 3.52×10-3 M SrH2.

Answer 1

Theory: We need to determine the resultant [OH-] Concentration by mixing both solution:

Step 1: Mg[OH]2 ----> Mg2+ + 2OH-

Volume = V1 = 5.83 mL = 5.83x10^-3 L

moles of Mg[OH]2 = molarity*V1 = 4.19x10^-2 *5.83x10^-3 = 3.574x10^-4 moles

moles of [OH]- = 2*mole of Mg[OH]2 = 2*3.574x10^-4 = 7.148x10^-4 moles

Step 2: SrH2 --: Sr2- + 2H+

Volume = V2 = 918 mL = 0.918  L

moles of SrH2 = molarity*V2 = 3.52x10^-3 *0.918 = 32.314x10^-4 moles

moles of [H]- = 2*mole of SrH2 = 2*32.314x10^-4 = 64.627x10^-4 moles

Step 3: [H] + [OH-] = H2O

limiting reactant is [OH-] as its moles are less than [H+]

so remaining mole of [H+] =64.627x10^-4 moles -  7.148x10^-4 moles =  57.479 x10^-4 moles

total volume of solution = V1 + V2 = 5.83+918 = = 923.83 ml = 0.92383 Liter

So [H+] = mole/Volume = 57.479 x10^-4 /0.92383 = 62.21822 x10^-4 M

Step 4: pH = -Log [H+]

= -Log[ 62.21822 x10^-4 ] = 2.206

pOH = 14-pH = 14-2.206 = 11.79