Question

Determine the pH of a solution when 26.6 mL of 0.029 M HNO3 is mixed with 16.7 mL of....

A) 0.0120 M NaOH. D) 0.144 M KOH

B) distilled water

C) 0.0060 M HCl

Answer 1

no of moles of

no of moles of HNO3 = 0.029M x 0.0266 L = 0.0007714 mol

Part A

no og moles of NaOH = 0.012 M x 0.0167 L = 0.0002004 mol

0.0002004 mol of NaOH will react with 0.0002004 mol of HNO3

so no of moles of HNO3 remainins = 0.0007714 mol - 0.0002004 mol = 0.000571 mol

total vulume 26.6 mL + 16.7 = 43.3 mL   = 0.0433 L

concentration of HNO3 remaining = 0.000571 mol / 0.0433 L   = 0.01319 M

pH = -log(H+) = -log (0.01319)

= 1.8798

= 1.9

Part B

no of moles of HNO3 = 0.029M x 0.0266 L = 0.0007714 mol

total vulume 26.6 mL + 16.7 = 43.3 mL   = 0.0433 L

concentration of HNO3 after dilution = 0.0007714 mol / 0.0433 L = 0.01781 M

pH = -log(0.01781) = 1.75

Part C

no of moles of HNO3 = 0.029M x 0.0266 L = 0.0007714 mol

no of moles of HCl = 0.006 M x 0.0167 L = 0.0001002 mol

total moles of H+ = 0.0007714 mol + 0.0001002 mol = 0.0008716 mol

total vulume 26.6 mL + 16.7 = 43.3 mL   = 0.0433 L

concentration of [H+] = 0.0008716 mol / 0.0433 L   = 0.02013 M

pH = -logH+ = -log(0.02013)   = 1.7

Part D

no of moles of HNO3 = 0.029M x 0.0266 L = 0.0007714 mol

no fo moles of KOH = 0.144 M x 0.0167L = 0.0024048 mol

one mole of HNO3 will react with one mole of KOH

no o fmoles of KOH remaining = 0.0024048 mol - 0.0007714 mol = 0.0016334 mol

total vulume 26.6 mL + 16.7 = 43.3 mL   = 0.0433 L

concentration of KOH remaining = 0.0016334 mol / 0.0433 L = 0.0377 M

pOH = -log(0.0377) = 1.42

pH = 14-pH = 14 - 1.42 = 12.57