sally wants to make a buffer solution with a pH of 5.66 with a total volume of 7
ID: 775494 • Letter: S
Question
sally wants to make a buffer solution with a pH of 5.66 with a total volume of 750ml, using aniline(C6H5NH2 -pKb=9.41)and its conjugate acid. First she adds 2.35g of C6H5NH3Cl (129.59g/mol), to about 200ml of water, in a 750ml flask to dissolve the salt. next she adds a volume of a 3.74M C6H5NH2 stockroom solution to the flask. finally she adds water until the volume of the solution is 750ml. What volume(v C6H5NH2 in ml) of the 0.28 M stockroom solution didshe end up using to get the 5.66 pH value?
Explanation / Answer
Since it is a buffer... we will work with the alternate form of Henderson-Haselbach equation:
pOH = pKb + log ([Conjugate]/[Base]
V total = 0.75 L
pH = 5.66
Remember that 14 = pH + pOH ==> pOH = 14- 5.66 = 8.34
pKb = 9.41
NOte
m = 2.35 g of aniline
MW aniline = 129.59 g/gmol
then moles of aniline = m/MW = 2.35g/129.59 g/gmol = 0.018 gmol of aniline
V water = 0.2 L
V of aniline = ?
M of aniline = 3.74 M
V water extra = ? not know, but must fill 750 ml
We will need to calculate first the amount of moles needed to get the concentratinon of [base] using:
pOH = pKb + log ([Conjugate]/[Base]
solve for base
8.34 = 9.41 + log ([Conjugate]/[Base]
We need to calculate the conjugate concentration
The final volume must be 750 ml or 0.75 L and we have 0.018 gmols so
M = moles / V = 0.018/0.75 =
Now substitute
8.34 = 9.41 + log ([0.024]/[Base])
0.08511 = 0.024 / [Base]
[Base] = 0.024/0.08511 = 0.2819
Now... we know the concentration and the moles needed... find out the volume!
M1V1 = M2V2 NOTE (m1 and v1 is molarity of the base, and volume we added first... the amount of moles is constant so the second molarity, the one calculated with the formula, and the second volume, the final total volume, must be equal)
3.74 * V1 = 0.2819*750 ml
V1 = 56.5 ml of base must be added
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.