3. Determine the pH (to two decimal places) of the solution that is produced by mixing...

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Question

3. Determine the pH (to two decimal places) of the solution that is produced by mixing 70.1 mL of 3.01x10-1 M HNO3 with 766 m

Answer 1

Answer #1

Since HNO3 is a strong acid and NaOH is a strong base, so it will dissociate completely into H+ and OH- ions

Number of moles of HNO3 = Volume of solution (in L) * Molarity (M) = 70.1/1000 * 0.301 = 0.0211 moles

Number of moles of NaOH = Volume of solution (in L) * Molarity (M) = 766/1000 * 0.0684 = 0.0523944 moles

Number of moles of excess NaOH = (0.0529344 - 0.0211) = 0.031834 moles

Total Volume = 70.1 + 766 = 836.1 ml

Molarity of OH- left = (Number of moles)/(Volume of solution (in L)) = 0.031834/0.8361 = 0.03807 M

pOH = -log[OH-] = -log(0.03807) = 1.42

pH = 14 - pOH = 14 - 1.42 = 12.58

Hence the pH will be equal to 12.58

Note - Post any doubts/queries in comments section.

Answer 2

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