A few Acid + Base reaction questions

[TrueTears]

Could some1 just check this working done by me, not sure if it is correct and book has no answers. Thanks !

1. A solution is produced by mixing 60mL of 0.20M KOH and 40mL of 0.10M HNO3.
a) calculate the number of mole of KOH
n=cV n = o.2 x (60/1000) = 0.012mole.
b) Hence determine the number of mole of OH- added.
mole OH- = 0.012
c) calculate the number of mole of HNO3
n=cV n=0.1 x (40/1000) = 0.004 mole
d) Hence determine the number of mole of H+ added.
0.004 mole
e) Use your previous answers to calculate the final [H+] AND the pH of the solution.
In KOH [OH-]= 0.2M therefore [H+] x [OH-]=10^-14 so [H+] = 5x10^-14 M
n = 5x10^-14 x (60/1000) = 3x10^-15 mole of H+ in KOH.
so 3x10^-15+0.004 = c x 0.1
c = 0.04 M
[H+] = 0.04 M and pH= -log([H+]) so pH = 1.4

Could some1 please check part a for this Q but for part b im just not sure how to approach it. (btw wat does it mean by 'required an average titre of hydrochloric acid of 16.30mL.'
Question 2: Ar ( H) = 1 Ar (Cl) = 35.5

A student proposes to determine the concentration of a solution of hydrochloric acid by reacting it with a standard solution of 0.100 M of sodium carbonate, Na2CO3.
a) write a balanced equation including states for the reaction
2HCL (aq) + Na2CO3 (aq) ---> 2NaCl (aq) + H2O (l) + CO2 (g)
b) if the student took 25.00 mL aliquots (samples) of the sodium carbonate solution and required an average titre of hydrochloric acid of 16.30 mL calculate
1. the molarity of the hydrochloric acid solution.
2. the mass of hydrogen chloride gas required to make 500 mL of the HCL solution
3. what substance was placed in the burette in the analysis.

any help would be greatly appreciated thanks!

 

[Pandemonium]

I don't know, I tend to not be able to read other people's writings, so please forgive me for doing it on my own.


1. A solution is produced by mixing 60mL of 0.20M KOH and 40mL of 0.10M HNO3.
a) calculate the number of mole of KOH
b) Hence determine the number of mole of OH- added.
c) calculate the number of mole of HNO3
d) Hence determine the number of mole of H+ added.
e) Use your previous answers to calculate the final [H+] AND the pH of the solution.

*a) n(KOH) = cv = 0.20 x 0.06 = 0.012 mol
*b) n(OH-) = n(KOH) = 0.012 mol
*c) n(HNO3) = cv = 0.10 x 0.04 = 0.004 mol
*d) n(H+) = n(HNO3) = 0.004 mol
*e) Acknowledge that according to the equation: H+(aq) + OH-(aq) -> H2O(l), there is a one-to-one relationship. According to the amount of mols that we have, there will be an excess of OH-.

n(OH-)remaining = n(OH-)initial - n(OH-)used               * n(OH-)used = n(H+)
n(OH-)remaining = 0.012 - 0.004 = 0.008 mol

Remember that final v is equal to the sum of the two volumes that were mixed.
Therefore, final v = 0.06 + 0.04 = 0.01

[OH-] = n/v = 0.008/(0.06+0.04) = 0.08 M

Acknowledge that Kw (power of water) has the value 10^-4 at 25 degrees C, and its relationship is:

Kw = [H3O+].[OH-]
10^-14 = [H3O+].(0.08)
[H3O+] = 0.000000000000125 M
pH = -log10[H3O+] = 12.90 = 13 (2 significant figures)

[Pandemonium]

Could some1 please check part a for this Q but for part b im just not sure how to approach it. (btw wat does it mean by 'required an average titre of hydrochloric acid of 16.30mL.'

What they mean is that (in the following question,) the 25.00mL solutions required an average of 16.30mL of solution to become fully neutralised. This is the value that you use in your calculations. (Three trials are used in order to minimalise random error)

Question 2:
A student proposes to determine the concentration of a solution of hydrochloric acid by reacting it with a standard solution of 0.100 M of sodium carbonate, Na2CO3.
a) write a balanced equation including states for the reaction
b) if the student took 25.00 mL aliquots (samples) of the sodium carbonate solution and required an average titre of hydrochloric acid of 16.30 mL calculate
1. the molarity of the hydrochloric acid solution.
2. the mass of hydrogen chloride gas required to make 500 mL of the HCL solution
3. what substance was placed in the burette in the analysis.

*a) 2HCl (aq) + Na2CO3 (aq) --> 2NaCl (aq) + H2O (l) + CO2 (g)
*b) i) First, draw a map or something about the things that we know for sure.
[Na2CO3], v(Na2CO3) -> therefore, we can determine n(Na2CO3)
Determining n(Na2CO3), we can calculate n(HCl) to neutralise it, and with v(HCl), we can find [HCl] which is what we want.

eg. [Na2CO3] + v(Na2CO3) -> n(Na2CO3) -> n(HCl) -> n(HCl) + v(HCl) -> [HCl]

n(Na2CO3) = c.v = 0.100 x 0.025 = 0.0025 mol

Recognise here that there is a 2:1 relationship regarding HCl and Na2CO3. Therefore, for every Na2CO3 reacted, we require 2 HCl to be consumed.

n(Na2CO3) = 2n(HCl)
n(HCl) = n(Na2CO3)/2 = 0.0025/2 = 0.00125 mol

Knowing n(HCl), we can now find [HCl] using the value we just found in conjunction with v(HCl).
According to the information, an average titre of 16.30mL was used. Use this figure.

[HCl] = n/v = 0.00125/0.01630 = 0.0767 M

ii) You might think that question requires the knowledge that HCl(g) dissolves in water to make HCl(aq). It doesn't. Weeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee.
However, what we need to know here is that we need a volume of 500mL of solution. Therefore our first priority is to, using this value, find the amount that exists in this solution. (Use the [HCl] value we determined before)

n(HCl) = cv = 0.0767 x 0.500 = 0.03834356 mol

The question states find the mass, so find it.

m(HCl) = n.M = 0.03834356 x (35.5+1) = 1.40 g

iii Well, this is basically... what is it? The burette contains the solution you are titrating with. This means it is the solution you are adding to the beaker. According to the question, HCl is being reacted with HCO3. So it's HCO3.

Also, the burette always contains something that has known precise concentration, so you can always choose that.

LOL. I just wanted to do some reading comprehension.

[TrueTears]

Could some1 please check part a for this Q but for part b im just not sure how to approach it. (btw wat does it mean by 'required an average titre of hydrochloric acid of 16.30mL.'

What they mean is that (in the following question,) the 25.00mL solutions required an average of 16.30mL of solution to become fully neutralised. This is the value that you use in your calculations. (Three trials are used in order to minimalise random error)

Question 2:
A student proposes to determine the concentration of a solution of hydrochloric acid by reacting it with a standard solution of 0.100 M of sodium carbonate, Na2CO3.
a) write a balanced equation including states for the reaction
b) if the student took 25.00 mL aliquots (samples) of the sodium carbonate solution and required an average titre of hydrochloric acid of 16.30 mL calculate
1. the molarity of the hydrochloric acid solution.
2. the mass of hydrogen chloride gas required to make 500 mL of the HCL solution
3. what substance was placed in the burette in the analysis.

*a) 2HCl (aq) + Na2CO3 (aq) --> 2NaCl (aq) + H2O (l) + CO2 (g)
*b) i) First, draw a map or something about the things that we know for sure.
[Na2CO3], v(Na2CO3) -> therefore, we can determine n(Na2CO3)
Determining n(Na2CO3), we can calculate n(HCl) to neutralise it, and with v(HCl), we can find [HCl] which is what we want.

eg. [Na2CO3] + v(Na2CO3) -> n(Na2CO3) -> n(HCl) -> n(HCl) + v(HCl) -> [HCl]

n(Na2CO3) = c.v = 0.100 x 0.025 = 0.0025 mol

Recognise here that there is a 2:1 relationship regarding HCl and Na2CO3. Therefore, for every Na2CO3 reacted, we require 2 HCl to be consumed.

n(Na2CO3) = 2n(HCl)
n(HCl) = n(Na2CO3)/2 = 0.0025/2 = 0.00125 mol

Knowing n(HCl), we can now find [HCl] using the value we just found in conjunction with v(HCl).
According to the information, an average titre of 16.30mL was used. Use this figure.

[HCl] = n/v = 0.00125/0.01630 = 0.0767 M

ii) You might think that question requires the knowledge that HCl(g) dissolves in water to make HCl(aq). It doesn't. Weeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee.
However, what we need to know here is that we need a volume of 500mL of solution. Therefore our first priority is to, using this value, find the amount that exists in this solution. (Use the [HCl] value we determined before)

n(HCl) = cv = 0.0767 x 0.500 = 0.03834356 mol

The question states find the mass, so find it.

m(HCl) = n.M = 0.03834356 x (35.5+1) = 1.40 g

iii Well, this is basically... what is it? The burette contains the solution you are titrating with. This means it is the solution you are adding to the beaker. According to the question, HCl is being reacted with HCO3. So it's HCO3.

Also, the burette always contains something that has known precise concentration, so you can always choose that.

LOL. I just wanted to do some reading comprehension.

ahh thanks for ur help but just for n(Na2CO3) = c.v = 0.100 x 0.025 = 0.0025 mol wudnt the c change since only 25mL of it was taken from the total amount of Na2CO3. Since the concentrate mol per litre if there was less Volume wouldnt the C decrease?

and also for Q 2 part iii where did HCO3 come from "HCl is being reacted with HCO3. So it's HCO3." but besides that yeah i get everything else

[Mao]

Pandemonium: good answers except for the last bit

you don't necessarily have to have a known concentration solution in the burette. The standard you prepare can be either in the burette or in the conical flask.
such as in this case, unknown concentration of HCl was added to an accurate volume of precisely concentrated Na2CO3 [the standard].
"if the student took 25.00 mL aliquots (samples) of the sodium carbonate solution and required an average titre of hydrochloric acid of 16.30 mL"

the chemical which the titre refers to is the chemical in the burette.
specifically, "aliquot" refers to the volume delivered by the pipette (accurate volume, in conical flask)
"titre" refers to the volume delivered by the burette (accurate volume, in the burette added to the conical flask)
hence why titration is called "volumetric" analysis.


so, the answer to 2 b iii) is unknown concentration of HCl

[Pandemonium]

Aw .
Anyway, concentration doesn't decrease when you take an aliquot.

[TrueTears]

I don't know, I tend to not be able to read other people's writings, so please forgive me for doing it on my own.


1. A solution is produced by mixing 60mL of 0.20M KOH and 40mL of 0.10M HNO3.
a) calculate the number of mole of KOH
b) Hence determine the number of mole of OH- added.
c) calculate the number of mole of HNO3
d) Hence determine the number of mole of H+ added.
e) Use your previous answers to calculate the final [H+] AND the pH of the solution.

*a) n(KOH) = cv = 0.20 x 0.06 = 0.012 mol
*b) n(OH-) = n(KOH) = 0.012 mol
*c) n(HNO3) = cv = 0.10 x 0.04 = 0.004 mol
*d) n(H+) = n(HNO3) = 0.004 mol
*e) Acknowledge that according to the equation: H+(aq) + OH-(aq) -> H2O(l), there is a one-to-one relationship. According to the amount of mols that we have, there will be an excess of OH-.

n(OH-)remaining = n(OH-)initial - n(OH-)used               * n(OH-)used = n(H+)
n(OH-)remaining = 0.012 - 0.004 = 0.008 mol

Remember that final v is equal to the sum of the two volumes that were mixed.
Therefore, final v = 0.06 + 0.04 = 0.01

[OH-] = n/v = 0.008/(0.06+0.04) = 0.08 M

Acknowledge that Kw (power of water) has the value 10^-4 at 25 degrees C, and its relationship is:

Kw = [H3O+].[OH-]
10^-14 = [H3O+].(0.08)
[H3O+] = 0.000000000000125 M
pH = -log10[H3O+] = 12.90 = 13 (2 significant figures)

What does this mean?
"*e) Acknowledge that according to the equation: H+(aq) + OH-(aq) -> H2O(l), there is a one-to-one relationship. According to the amount of mols that we have, there will be an excess of OH-.

n(OH-)remaining = n(OH-)initial - n(OH-)used               * n(OH-)used = n(H+)
n(OH-)remaining = 0.012 - 0.004 = 0.008 mol
why does n(OH-)used = n(H+) ?

to find the [H+] dont u just find the mole of H+ in KOH and the mole of H+ in HNO3 then divide by the total volume which is 0.1 and u got the [H+]?

[Mao]

no.

in KOH, you have two species: K+ (spectator ion) and OH- (base)
in HNO3, you also have two species: H+ (acid) and NO3- (spectator ion)

acid reacts with base, OH-(aq) + H+(aq) -> H2O(l)
i.e. if you have one mole of OH- and one mole of H+ and you mix them together, they react to form one mole of H2O and you won't have any OH- or H+ left.

applying it to this case, you have 0.012 mol of OH-, and 0.004 mol of H+. Here, OH- is in excess, so 0.004 mol of H+ react with 0.004 mol of OH-, and the left-overs remain, 0.008 mol of OH-

[TrueTears]

ah yeah thanks i understand. and we just did the titration prac today. Makes the other Q much clearer

and for this
Recognise here that there is a 2:1 relationship regarding HCl and Na2CO3. Therefore, for every Na2CO3 reacted, we require 2 HCl to be consumed.

n(Na2CO3) = 2n(HCl)
n(HCl) = n(Na2CO3)/2 = 0.0025/2 = 0.00125 mol

shouldnt it be 2 x 0.0025. Because every Na2CO3 reacted there would be double the amount of HCl used. so its (2/1) x 0.0025 = 0.005 mole of HCl?

[Pandemonium]

Uhh... yes that would be my error.
I'm not going to blame Unit 3 being a long time ago, but I just did.

Sorry bout that -- but the method is correct so you can do it yourself now.

[TrueTears]

yeap thanks heaps ^^ understand it fully