O ACIDS AND BASES Calculating the pH of a weak acid titrated with a strong base

Question

O ACIDS AND BASES Calculating the pH of a weak acid titrated with a strong base An analytical chemist is titrating 143.7 mL of a 0.9800 M solution of hydrazoic acid (HN3) with a 0.5600 M solution of KOH. The pK, of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 264.8 mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added Round your answer to 2 decimal places. PH

Explanation / Answer

HN3 = 143.7 ml of 0.9800M

number of moles of HN3= 0.9800Mx.143.7L=0.1408 moles

KOH= 264.8ml of 0.5600M

number of moles of KOH= 0.5600Mx0.2648L=0.1483 moles

number of moles of KOH is greaterthan HN3 acid

so the nature of th esolution is base

buffer solution is not possible.

remaining number of moles of KOH= 0.1483 - 0.1408 =0.0075 moles

total volume = 143.7+264.8 =408.5ml = 0.4085L

concentrtio of OH- = 0.0075/0.4085 =0.0184M

[OH-]= 0.0184M

-log[OH-]= -log(0.0184)

POH= 1.735

PH+POH=14

PH=14-POH

PH= 14- 1.735

PH= 12.26

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