How do you do this? 25 mL of 0.2 M NH3 (Kb = 1.8e-5) + 10mL of 0.2 M HCl. What is the pH of solution?

Question

1) 25 mL of 0.2 M NH3 (Kb = 1.8e-5) + 10mL of 0.2 M HCl. What is the pH of solution?

I got the new concentrations: .143 M NH3; .057 M HCl

But how do i neutralize the reaction before proceeding on?

is it: NH3 + H => NH4 ?

I'm not sure. I'm stuck.

2) Calculate pH of 25 mL of 0.2 M CH3COOH + 25 mL 0.2 M NaOH.

Now on this one, i got the new concentrations but they were both 0.1 M. Do i have to neutralize the reaction/solution?

If they are both with equal solutions doesn't that make it an ideal buffer which means the pH=pKa?

Help me out plz.

Answer 1

1) 25 mL of 0.2 M NH3 (Kb = 1.8e-5) + 10mL of 0.2 M HCl. find moles:

0.025 L of 0.2 mol/litre NH3 = 0.0050 moles NH3

0.010 L of 0.2 mol/litre HCl = 0.0020 moles H+

these two react, & then you have

0.0020 moles NH4+ produced

0.0030 moles NH3 left over

NH3 in water -->  ---> -->    NH4+   &   OH-

(0.003mol/0.035 L)         (0.002mol/0.035 L)   & OH-

     (0.0857 molar    --->   (0.0571 molar)     &  OH-

Kb = [NH4+] [OH-]  / [NH3]

1.8e-5 =  [0.0571[OH-]  / [0.0857]

OH- = 2.7e-5

pOH = 4.57

14 - pOH = pH

your answer: pH = 9.43

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2) Calculate pH of 25 mL of 0.2 M CH3COOH + 25 mL 0.2 M NaOH.

moles :

0.025Litres @ 0.2mol/litre = 0.005 moles of acid & base=

make 0.005 moles of sodium acetate & water

CH3COO -  in water --> CH3COOH   &  OH-

0.005 mol/ 0.05Litre) -->   x    &   ....    ....   x

(0.10 molar ) ----->   x   &    x

this is the conjugate base of acetic acid, its Kb = Kw / Ka

Kb = 1e-14 / 1.8e-5 = 5.56e-10

Kb = [CH3COOH ] [OH-] / [CH3COO-]

5.56e-10 = [x ] [x] / [0.1]

x = OH- = 7.46e-6

pOH = 5.13

your answer is:pH = 8.87

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If they are both with equal solutions doesn't that make it an ideal buffer which means the pH=pKa?

pH=pKa when 1/2 of the acid is still there & 1/2 has been neutralized.  it would have been from mixing:

25 mL of 0.2 M CH3COOH + 12.5 mL 0.2 M NaOH

giving you 0.05 molar acid & 0.05 molar sodium acetate