# chemistry

Calculate the pH after 0.014 mole of gaseous HCl is added to 227.0 mL of each of the following buffered solutions. (Assume that all solutions are at 25°C.)

(a) 0.089 M NH3 and 0.17 M NH4Cl
(b) 0.74 M NH3 and 1.25 M NH4Cl

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1. I like to work in millimols.
millimoles NH3 = mL x M = 227.0 mL x 0.089 = approx 20 but you need a better number on this and all other estimates that follow.
millimoes NH4Cl = 227.0 x 0.17 = about 39

...................NH3 + HCl ==> NH4Cl
I..................20........0...............39
C............-...14......-14...............+14
E...................6........0.................53..

The problem doesn't give you the volume so use millimoles as the same as M. They aren't, of course, however, in the calculation the total volume cancels so it isn't necessary. Plut millimoles into the Henderson-Hasselbalch equation and solve for pH.

The second part is done the same way.

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2. B+ H2O <----->BH +OH^-

Kb=[BH][OH^-]/[B]

Kb=([BH]/[B])*[OH^-]

-logKb=-log([BH]/[B])*-log[OH^-]

pKb=-log([BH]/[B])+pOH

Variation of the Henderson-Hasselbach Equation:

pOH=pKb+log[BH]/[B]

14-pOH=pH

Convert to moles for each:

0.089 M NH3*0.227L= 0.020203 moles of NH3
0.17 M*0.227L= 0.03859 moles of NH4Cl

Addition of a strong acid to a weak base in its conjugate acid will increase the amount of conjugate acid and decrease the amount of weak base by the same amounts. You can use an ICE chart if you like, which can be beneficial in visualizing what is happening.

B + HCl ---> BH +Cl

BH=NH3Cl=0.03859+0.014=0.05259
B=NH3=0.020203-0.014=0.006203

Plug into equation:

let pKb=4.75 but use the one from your text.

pOH=4.75+log[0.05259]/[0.006203]

pOH=4.75+0.9283=5.678

pH=14-5.678=8.32

B.) Is the same way.

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