Question

# Calculate the pH after 5.00 mL of 0.60 M KOH is added to a 25.0 mL...

Calculate the pH after 5.00 mL of 0.60 M KOH is added to a 25.0 mL of a 0.55 M NH3/0.33 M NH4Cl buffer solution.

I'm completly lost on how to solve this.....

mol of KOH added = 0.6M *5.0 mL = 3.0 mmol

NH4+ will react with OH- to form NH3

Before Reaction:

mol of NH3 = 0.55 M *25.0 mL

mol of NH3 = 13.75 mmol

mol of NH4+ = 0.33 M *25.0 mL

mol of NH4+ = 8.25 mmol

after reaction,

mol of NH3 = mol present initially + mol added

mol of NH3 = (13.75 + 0.0) mmol

mol of NH3 = 13.75 mmol

mol of NH4+ = mol present initially - mol added

mol of NH4+ = (8.25 - 0.0) mmol

mol of NH4+ = 8.25 mmol

since volume is both in numerator and denominator, we can use mol instead of concentration

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.7447

we have below equation to be used:

This is Henderson–Hasselbalch equation

pOH = pKb + log {[conjugate acid]/[base]}

= 4.7447+ log {8.25/13.75}

= 4.52

we have below equation to be used:

PH = 14 - pOH

= 14 - 4.5229

= 9.48

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