acid and base calculation question help

[allthebest]

Please help me with this question:

Question:

25.0 cm3 of 1.00 × 10–2 mol dm–3 hydrochloric acid solution is added to 50.0 cm3 of
1.00 × 10–2 mol dm–3 aqueous ammonia solution. Calculate the concentrations of both ammonia and ammonium ions in the resulting solution and hence determine the pH of the solution.

Mark scheme:

initial amount of HCl =  × 1.00 × 10–2 = 2.50 × 10–4 mol
and initial amount of NH3 =  × 1.00 × 10–2 = 5.00 × 10–4 mol;
final amount of NH4+ and NH3 both = 2.50 × 10–4 mol;
final [NH4+] and [NH3] both =  = 3.33 × 10–3 mol dm–3;
[OH–] = Kb ×  = Kb = 10–4.75 /1.78 × 10–5;
pOH = 4.75 hence pH = 9.25;

I don't quite understand the reactions involving ammonia solution, like are both NH4+ and NH3 are present there? Why do we have to know the final amount of them both and where did it come from?

Thank you

[Msj Chem]

"I don't quite understand the reactions involving ammonia solution, like are both NH4+ and NH3 are present there?" 

HCl reacts with NH₃ to produce NH₄Cl

HCl + NH₃ ==> NH₄Cl

The NH₃ is the excess reactant which is why you have some left over. 

The NH₄⁺ comes from the dissociation of the salt NH₄Cl (assume that the salt fully dissociates). 

"Why do we have to know the final amount of them both and where did it come from?"

You have equal final concentrations of NH₃ and NH₄⁺ which acts as a buffer solution.

Subtract the amount in mol that reacted from the initial amount in mol. Use the equation C=n/V to get the concentration.  

Use the Henderson-Hasselbach equation (like in the previous example you posted) to calculate the pH and pOH. 

 

 

Edited November 30, 2016 by Msj Chem

[allthebest]

19 hours ago, Msj Chem said:

"I don't quite understand the reactions involving ammonia solution, like are both NH4+ and NH3 are present there?" 

HCl reacts with NH₃ to produce NH₄Cl

HCl + NH₃ ==> NH₄Cl

The NH₃ is the excess reactant which is why you have some left over. 

The NH₄⁺ comes from the dissociation of the salt NH₄Cl (assume that the salt fully dissociates). 

"Why do we have to know the final amount of them both and where did it come from?"

You have equal final concentrations of NH₃ and NH₄⁺ which acts as a buffer solution.

Subtract the amount in mol that reacted from the initial amount in mol. Use the equation C=n/V to get the concentration.  

Use the Henderson-Hasselbach equation (like in the previous example you posted) to calculate the pH and pOH. 

 

 

Thank you!!

May I ask another question..?

The question is:

Ammonia, NH3 , can be used to clean ovens. The concentration of hydroxide ions,
OH– (aq), in a solution of ammonia is 3.98 × 10–3 mol dm–3. Calculate its pH, correct to
one decimal place, at 298 K.

and mark scheme:

 

[H O ] (1.00 10 ) 2.51 10 (moldm )
[OH ] (3.98 10 )
K −
+ − −
− −
= = × = ×
×
;
( 12 )
3 pH = −log[H O+ ] = −log(2.51×10− ) =11.6 ; [2]
OR
pOH = (−log (5.98×10−3 ) =)2.4 ;
pH = (14.00 − 2.40) =11.6 ;

 

Why can't we directly do pOH= -log(3.98 x 10^-3) and then do 14-pOH?

Why do we need to use the Kw or why is pOH = (−log (5.98×10−3 )..??

I'd be very thankful if you would answer this questions also. Thank you.

 

[Msj Chem]

There are two ways to calculate this.

One way is to use the Kw to find the [H⁺], and then use pH = -log [H⁺]

The second way is to use pOH = -log[OH^_-]

There is a mistake in the mark scheme, it should read -log (3.98 x10^-3) = 2.4,