Question

500ml of a buffer solution contains 0.050 mol nahso3 and 0.031

molna2so3

a) what is the ph of the solution?

b)write the net ionic equation for the reaction whne naoh is
addedto this buffer.

c) calculate the new ph after 10ml of 1.0 m naoh is added to
thebuffer solution.

d) calculate the new ph after 10ml of 1.0 m naoh is added to
500mlof pure water.

E) explain why the ph of water changed so much as compared to theph
of the buffer.

Answer 1

The answers are explained below

Step-by-step explanation:

a)

Given: concentration of salt/base = 0.031

concentration of acid = 0.050

we have

PH = PK a + log[salt]/[acid] = 1.8 + log(0.031/0.050) = 1.59

b)

we have HSO₃⁻ + OH⁻ ------> SO₃²⁻ + H₂O

Moles i............0.05...................0.01.................0.031.....................0

Moles r...........-0.01.................-0.01................0.01........................0.01

moles f...........0.04....................0....................0.041.....................0.01

c)

we will use the first equation but substituting concentration of base as 0.031 + 10ml = 0.031 + 0.010 = 0.041

Hence, we have

PH = PK a + log[salt]/[acid] = 1.8 + log(0.041/0.050) = 1.71

d)

pOH = -log (0.01/0.510) = 1.71

pH = 14 - 1.71 = 12.29

e)

Because the buffer solution (NaHSO3-Na2SO3) can regulate pH changes. when a buffer is added to water, the first change that occurs is that the water pH becomes constant. Thus, acids or bases (alkali = bases) Additional may not have any effect on the water, as this always will stabilize immediately.