6. Aniline, C&HsNH2;, is used in the manufacture of some perfumes. What is the p

Question

6. Aniline, C&HsNH2;, is used in the manufacture of some perfumes. What is the pH of a 0.035 M solution of aniline at 25 C Ko= 4.2 x 10-10 at 25°C Ammonium nitrate, NH&NOs;, is administered as an intravenous solution to 7. patients whose blood pH has deviated from the normal value of 7.40. Would this substance be used for acidosis (blood pH 7.40)? What is K'v for the F ion, the ion added to the publie water supply to protect 8. teeth? For HF, K,-6.8 × 10". Household bleach is a 5% solution of sodium hypochlorite, NaClO. This corresponds to a molar concentration of about 0.70 M NaCIO 9. What is the [OH 1 and the plt of the solution? 4,5X1G For HCIO, K-3.5x10- 10. Calculate the pH of a 0.575 M of sodium acetate. What is the pH of a buffer made by mixing 1.00 L of 0.020 M benzoic acid, HC,Hs02. with 3.00 L of 0.060 M sodium benzoate, NaC,HsO2? Ka for benzoic acid is 6.3 × 10-5 What is the [H O'] for a buffer solution that is 0.250 M in acid and 0.600 M in the corresponding salt if the weak acid K-5.80 x 10 12. Calculate the pH of a solution containing 0.22 M acetic acid and 0.13 M sodium acetate. (Kaofacetic acid= 1.8× 10-5) 13. 45 14. Determine the number of moles of sodium acetate that must be added to 250.0 mLof0.16 M acetic acid in order to prepare a pH 4.70 buffer (K.-1.76x10,s).

Explanation / Answer

Q6

This is a base in water so, let the base be CH3NH2 = "B" and CH3NH3+ = HB+ the protonated base "HB+"

there are free OH- ions so, expect a basic pH

B + H2O <-> HB+ + OH-

The equilibrium Kb:

Kb = [HB+][OH-]/[B]

initially:

[HB+] = 0

[OH-] = 0

[B] = M

the change

[HB+] = x

[OH-] = x

[B] = - x

in equilibrium

[HB+] = 0 + x

[OH-] = 0 + x

[B] = M - x

Now substitute in Kb

Kb = [HB+][OH-]/[B]

Kb = x*x/(M-x)

x^2 + Kbx - M*Kb = 0

x^2 + (4.2*10^-10)x - (0.035)(4.2*10^-10) = 0

solve for x

x = 3.833*10^-6

substitute:

[HB+] = 0 + x =  3.833*10^-6 M

[OH-] = 0 + x =  3.833*10^-6 M

[B] = M - x = 0.035 - 3.833*10^-6= 0.0349 M

pH = 14 + pOH = 14 + log(3.833*10^-6) = 8.583

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