Question

1) What is the pH of a 0.246 M aqueous solution of sodium acetate, NaCH3COO? This solution is acidic, basic or neutral?

2) The substance benzoic acid (C₆H₅COOH) is a weak acid (K_a = 6.3×10^-5). What is the pH of a 0.110 M aqueous solution of sodium benzoate, NaC₆H₅COO? This solution is acidic, basic or neutral?

3) What is the pH of a 9.17×10^-2 M aqueous solution of ammonium nitrate, NH₄NO₃? This solution is acidic, basic or neutral?

Answer 1

AS we know that Sodium acetate is a strong salt
CH3COONa --------> CH3COO- + Na+ ([CH3COO-]= 0.1 )



the equillibrium is
CH3COO- + H2O <------> CH3COOH + OH-
initial 0.1 0 0
change -x x x
at equilibrium 0.1 -x x x

Kb = [CH3COOH] [OH-] /[CH3COO-]
Ka for acetic acid = 1.75 × 10 ⁻⁵

Kb for acetate = Kw/Ka =10-14 /1.75 × 10 ⁻⁵ = 5.71 * 10^-10

5.71 * 10^-10    = X2 / 0.1 -x

0.571 * 10^-10 -  5.71 * 10^-10 x = x²

x² +   5.71 * 10^-10 x - 0.571 * 10^-10 =0

( using quadratic formulae we solve for x)

x= -b +_ b²- 4ac /2a

= - 5.71 * 10^-10   +_   (5.71 * 10^-10)²   - 4* (1) (0.571 * 10^-10) /2(1)

=  - 5.71 * 10^-10  +_   32.60 * 10^-20 - 2.284 * 10^-10 /2 ( neglecting 32.60 * 10^-20 as it is a very small value compare to 2.284 * 10^-10 )

= - 5.71 * 10^-10  +1.511 * 10^-5 /2    ( neglecting 5.71 * 10^-10 as it is a very small value comparison to1.511 * 10^-5 )

x = 0.755 * 10^{-5 =} [OH^_]

pOH = - log (  0.755 * 10^-5) = 5.12

as pOH + pH= 14 or  pH = 14 - pOh= 14 - 5.12 = 8.88

the solution is basic .

2) as we know that  CH3COO- is the conjugate base of CH3COOH.

CH3COO- + H2O <-------->CH3COOH + OH-

Kb = Kw / Ka = 1 x 10^-14 /6.3 x 10^-5 = 1.58 * 10 ^-10

the equillibrium is
CH3COO- + H2O <------> CH3COOH + OH-
initial 0.110 0 0
change -x x x
at equilibrium 0.11 -x x x

Kb = [CH3COOH][OH-] / [CH3COO-]   = (x)(x) / (0.11 - x)
1.58 * 10 ^-10 = x² / 0.11-x

0.17  * 10 ^-10 -  1.58 * 10 ^-10 x =  x²

x² +  1.58 * 10 ^-10 x - 0.17  * 10 ^-10=0

( using quadratic formulae we solve for x)

x= -b +_ b²- 4ac /2a

= -1.58 * 10 ^-10  +_   (1.58 * 10 ^-10)²   - 4* (1) (0.17  * 10 ^-10) /2(1)

=  -1.58 * 10 ^-10  +_    - 0.68 * 10^-10 /2 ( neglecting   2.49 * 10^-20 as it is a very small value    compare to 2.284 * 10^-10  )

=   -1.58 * 10 ^-10 +  0.82 * 10^-5 /2 ( neglecting 1.58 * 10 ^-10   as it is a very small value comparison to 0.82 * 10^-5    )

x = 0.41 * 10^{-5 =} [OH^_]

pOH = - log (  0.41 * 10^-5) = 5.38

as pOH + pH= 14 or  pH = 14 - pOh= 14 - 5.38 = 8.62

the solution is basic .

3)NH4NO3 ------> NH4+ + NO3-   

9.17 * 10-2

NH4+ + H2O <--------->   NH4 OH + H+ ( conjugate acid )

initial 9.17 * 10-2     0 0

equillibrium 9.17 * 10-2 -x x x

Ka = 10 ^-14 / 1.8 * 10^-5 = 5.55 * 10^-10

Ka= [NH4OH][H+]/[NH4+]

5.55 * 10^-10 = x2 /   9.17 * 10^-2 -x

50.89 * 10^{-12 -} 5.55 * 10^-10 x = x² or    x² + 5.55 * 10^-10 x - 50.89 * 10^-12 = 0

using quadratic equation

x= -b +_ b²- 4ac /2a

x= - 5.55 * 10^-10 +_   (5.55 * 10^-10 )2 - 4 (1) 50.89 * 10^-12 / 2(1)

=  - 5.55 * 10^-10 +_   (30.80 10^-20 - 203.56 * 10^-12 / 2(1) ( neglecting 30.80 10^-20 in compare to

203.56 * 10^{-12 )}

=    - 5.55 * 10^-10 +_    203.56 * 10^-12 / 2(1)       

=  - 5.55 * 10^-10 +_ 14.26 * 10^-6    /2 ( neglecting 5.55 * 10^-10 in comparison to 14.26 * 10^-6 )

= 7.13 *  10^-6 = H+

pH= - log H+ = - log 7.13 *  10^-6 = 5.14

solution is acidic