A buffer that contains 0.14 M of a base, B and 0.34 M of its conjugate acid BH +

Question

A buffer that contains 0.14 M of a base, B and 0.34 M of its conjugate acid BH+, has a pH of 9.49. What is the pH after 0.02 mol of Ba(OH)2are added to 0.52 L of the solution?

The answer is 9.79.

There is a worked out answer to this question already on chegg but It is hard for me to follow. One it is an image and it's a little blurry and small. Another reason I do not follow is because I do not see where they get the 0.4 that they use to subtract from the conjugate acid and add to the concentration of the base.

Thank you

Explanation / Answer

Solution :-

base = 0.14 M

conjugate acid = 0.34 M

so lets first calculate the pka of the acid

pH= pka + log [base / acid ]

9.49 = pka + log [0.14/0.34]

9.49 = pka + (-0.3853)

pka = 9.49-(-0.3853)

pka = 9.875

Now We have to find the pH after adding 0.02 mol Ba(OH)2

1 mol Ba(OH)2 have 2 mol OH-

therefore 0.02 mol Ba(OH)2 * 2 mol OH- / 1 mol Ba(OH)2 = 0.04 mol OH-

so we are adding 0.04 mol OH-

After adding the 0.04 mol OH-

OH- will react with acid and form the base

initial moles of acid = molarity * volume

                             = 0.34 mol per L * 0.52 L = 0.1768 mol

initial moles of base = 0.14 mol per L * 0.52 L = 0.0728 mol

after the reaction

mole of acid remain = 0.1768 mol - 0.04 mol = 0.1368 mol

moles of base after the reaction = 0.0728 mol + 0.04 mol = 0.1128 mol

volume remains the same

thereofre now lets calculate the pH using the Henderson equation

pH= pka + log ([base]/[acid])

pH= 9.875 + log [0.1128/0.1368]

pH= 9.875 + (-0.084)

pH= 9.79

So the pH after adding Ba(OH)2 is 9.79

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