Question

1) A solution contains 0.353 M potassium nitrite and 0.500 M nitrous acid.  

The pH of this solution is ___________ .

2) A solution contains 0.230 M sodium acetate and 0.410 M acetic acid.  

The pH of this solution is ___________ .

3) A solution contains 0.500 M NH₄Br and 0.413 M ammonia.  

The pH of this solution is____________ .

4) The compound ethylamine is a weak base like ammonia. A solution contains 0.189 M C₂H₅NH₃⁺ and 0.135 M ethylamine, C₂H₅NH₂.  

The pH of this solution is __________ .

5) A buffer solution is 0.500 M in HNO₂ and 0.335 M in NaNO₂. If K_a for HNO₂ is 4.5×10^-4, what is the pH of this buffer solution?

Answer 1

Question 1.

find pH of KNO2 and HNO2

this is a buffer

A buffer is any type of substance that will resist pH change when H+ or OH- is added.

This is typically achieved with equilibrium equations. Both type of buffer will resist both type of additions.

When a weak acid and its conjugate base are added, they will form a buffer

The equations:

The Weak acid equilibrium:

HA(aq) <-> H+(aq) + A-(aq)

Weak acid = HA(aq)

Conjugate base = A-(aq)

Neutralization of H+ ions:

A-(aq) + H+(aq) <-> HA(aq); in this case, HA is formed, H+ is neutralized as well as A-, the conjugate

Neutralization of OH- ions:

HA(aq) + OH-(aq) <-> H2O(l) + A-(aq) ; in this case; A- is formed, OH- is neutralized as well as HA.

Apply Henderson hasselbach equations

pH = pKa + log(KNO2 / HNO2)

pKa = 3.39 for HNO2

substitute

pH = 3.39 + log(0.353/0.50) = 3.2388

Question 2.

NaAcetate = 0.23 M, acetic acid = 0.41 M,

this is oncea gain, a buffer, use henderson hasselback equation

pH= pKa + log(NaAcetate / Acetic Acid)

pKa = 4.74 for acetic acid

substitute values

pH = 4.74 + log(0.23/0.41)

pH = 4.48894

Question 3.

there is NH4+ and NH3 (ammonia)

this is a buffer, once again

pH = pKa+ log(NH3/NH4+)

the acid is NH4+ (from NH4Br) and the base is NH3

pKa = 14-pKb = 14-4.75 = 9.25

substitute known data

pH = 9.25 + log(NH3/NH4Br) = 9.25 + log(0.413/0.5)

pH = 9.16698

this is basic, since this is ammonia buffer

Question 4.

similar as before,

the acid is C2H5NH3+, which can donate H+ to solution

the conjugate is C2H5NH2, therefore, the base

pH = pKa + log(C2H5NH2/C2H5NH3+)

pKa for ehtylmaine: pKa = 14-pKb = 14-3.2= 10.8

now, substitute

pH = 10.8 + log(0.135/0.189)

pH = 10.6538

Question 5.

this is a buffer, similar as in a and b

pH = pKa + log(NaNO2 / HNO2)

pKa = -log(Ka) = -log(4.5*10^-4) = 3.3467

pH= 3.3467 + log(0.335/0.5)

ph = 3.17277