Question
How many grams of NH4Cl must be added to 0.560 L of 1.90 M NH3 solution to yield a buffer solution with a pH of 8.50? Assume no volume change. Kb for NH3 = 1.8 10-5.
i am confused as how to find the grams?
i am confused as how to find the grams?
Answers
Devron
Let B= NH3 and BH=NH4Cl
pH+pOH=14
so, 14-8.50=pOH
and
pkb=-logkb
You will need to use a version of the henderson-hasselbalch equation.
pOH=pkb +log[BH/B]
Solve for the ratio,
So, [BH/B]= 0.755
0.560L*(1.90M)= moles of NH3
Substituting, you should have
0.775=BH/moles of NH3
Solve for BH
moles of BH*(53.49 g/mole)= mass of NH4Cl
pH+pOH=14
so, 14-8.50=pOH
and
pkb=-logkb
You will need to use a version of the henderson-hasselbalch equation.
pOH=pkb +log[BH/B]
Solve for the ratio,
So, [BH/B]= 0.755
0.560L*(1.90M)= moles of NH3
Substituting, you should have
0.775=BH/moles of NH3
Solve for BH
moles of BH*(53.49 g/mole)= mass of NH4Cl
DrBob222
I should point out here that you do not need to modify the Henderson-Hasselbalch equation.
pH = pKa + log (base)/(acid) is valid for acids and bases. In this case we calculate pKb from Kb, then
14-pKb = pKa, then proceed with the HH equation as usual.
8.50 = 9.25 + log (1.90/x)
Solve for x which is molar NH4Cl and convert to grams in 0.560 L.
pH = pKa + log (base)/(acid) is valid for acids and bases. In this case we calculate pKb from Kb, then
14-pKb = pKa, then proceed with the HH equation as usual.
8.50 = 9.25 + log (1.90/x)
Solve for x which is molar NH4Cl and convert to grams in 0.560 L.
bea
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