Use the Henderson-Hasselbalch equation to calculate the pH of a buffer that is 0

Question

Use the Henderson-Hasselbalch equation to calculate the pH of a buffer that is 0.158 M HClO and 0.099 M NaClO. The K_b for HCIO is 3.45 times 10^-7. NH_3 can react directly with BF_3, forming NH_3-BF_3. In this reaction. A) the NH_3 acts as a Bronsted base, accepting a proton from the BF_3 molecule B) the NH_3 acts as a Lewis base, donating a proton to the BF_3 molecule C) the NH_3 acts as a Bronsted acid, donating a proton to the BF_3 molecule D) BF_3 molecule acts as a Lewis acid, accepting an electron pair from the NH_3 molecule to form a coordinate covalent bond E) the BF_3 molecule acts as a Lewis base, donating an electron pair to the NH_3 molecule to form a coordinate covalent bond Put the following species in order of the most acidic substance? A) HClO_4 B) HClO C) Cl^- D) HClO_3 E) HClO_2 Put the following four species in order of the least acidic substance? A) HBr B) HI C) HF D) HCl Put the following four species in order of the most acidic substance? A) H_2 Se B) H_2 Te C) H_2 O D) H_2 S

Explanation / Answer

5) We have [HClO] = 0.158 M
[NaClO] = 0.099 M
Kb = 3.45 x 10-7
                       ClO- + H2O <--->    HClO +    OH-

Initial              0.099M                  0.158               0
Change           -x                      x           x
equilibrium       0.099M-x           0.158+x       x

Kb = [HClO][OH-]/[NaClO] = (0.158+x)x/(0.099 -x)
(since, x is very small we can take 0.158+x ~ 0.158; and 0.099-x ~ 0.099)
3.45 x 10-7 = 0.158x/0.099
5.52 x10-7 = x

[OH-] = 5.52 x 10-7
pOH = 6.26

pH = 14 -pOH = 14- 6.26 = 7.74

6) Answer: D
There is no proton transfer, so it is not a Bronsted acid-base reaction.
The lone pair on NH3 enters a vacant orbital on BF3 to give a new covalent bond.
the BF3 molecule acts as a Lewis acid, accepting an electron pair from the NH3
molecule to form a coordinate covalent bond.

7) Cl-<HClO<HClO2<HClO3<HClO4

HClO4 is the most acidic in the above

8)HF>HCl>HBr>HI

HF is most acidic in the above and least acidic is HI

9)H2O < H2S < H2Se < H2Te

H2Te is the most acidic in the above species.

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