The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

pH calculations

Hi guys!

I was wondering if you were able to help me with these questions?
Calculate the pH if the following weak acids:
a) 0.10 mol dm^3 NH4Cl
pKa (NH4+)= 9.25
Calculate the pH of the solution formed when:
a) 20cm^3 of 0.10moldm-3 HCl is added to 30cm^3 of 0.04moldm-3 NaOH
b)100cm^3 of water is added to 25cm^3 of 0.50moldm-3 KOH
c) 15cm^3 of 0.5moldm-3 H2S04 is added to 100cm^3 of 0.75moldm-3 KOH

Any help would be appreciated :smile:
Thank you!
Reply 1
Avatar for amnahaa
amnahaa
OP
9
bumppp :smile:
Calculate the pH of the solution formed when:
a) 20cm^3 of 0.10moldm-3 HCl is added to 30cm^3 of 0.04moldm-3 NaOH

1. Write a balanced equation for the reaction between HCl and NaOH
2. Calculate the moles of each one
3. Figure out which moles are in excess
4. Convert the excess moles back to concentration (using the NEW volume of 50 cm3)
5. If HCl is in excess - the molar ratio of HCl to H+ is 1:1, so find pH by simply substituting HCl concentration in the pH = -log[H+] equation
6. If NaOH is in excess, use the Kw expression to find H+ and then find pH. OH- conc = NaOH conc (1:1 ratio)


a) 0.10 mol dm^3 NH4Cl, pKa (NH4+)= 9.25

1. Convert pKa back to Ka (=10-pKa)
2. Use Ka = [H+]2/[HA] expression
3. [HA] = 0.10
4. Rearrange to find [H+]2, then square root to find [H+]
5. Use this value to find pH
Reply 3
Avatar for amnahaa
amnahaa
OP
9
Original post by flowerscat
Calculate the pH of the solution formed when:
a) 20cm^3 of 0.10moldm-3 HCl is added to 30cm^3 of 0.04moldm-3 NaOH

1. Write a balanced equation for the reaction between HCl and NaOH
2. Calculate the moles of each one
3. Figure out which moles are in excess
4. Convert the excess moles back to concentration (using the NEW volume of 50 cm3)
5. If HCl is in excess - the molar ratio of HCl to H+ is 1:1, so find pH by simply substituting HCl concentration in the pH = -log[H+] equation
6. If NaOH is in excess, use the Kw expression to find H+ and then find pH. OH- conc = NaOH conc (1:1 ratio)


a) 0.10 mol dm^3 NH4Cl, pKa (NH4+)= 9.25

1. Convert pKa back to Ka (=10-pKa)
2. Use Ka = [H+]2/[HA] expression
3. [HA] = 0.10
4. Rearrange to find [H+]2, then square root to find [H+]
5. Use this value to find pH

Thank you so much youre a life saver !
Original post by amnahaa
Hi guys!

I was wondering if you were able to help me with these questions?
Calculate the pH if the following weak acids:
a) 0.10 mol dm^3 NH4Cl
pKa (NH4+)= 9.25
Calculate the pH of the solution formed when:
a) 20cm^3 of 0.10moldm-3 HCl is added to 30cm^3 of 0.04moldm-3 NaOH
b)100cm^3 of water is added to 25cm^3 of 0.50moldm-3 KOH
c) 15cm^3 of 0.5moldm-3 H2S04 is added to 100cm^3 of 0.75moldm-3 KOH

Any help would be appreciated :smile:
Thank you!


is this a gcse question?
Reply 5
Avatar for amnahaa
amnahaa
OP
9
A level :smile: Don't worry!
Reply 6
Avatar for amnahaa
amnahaa
OP
9
Original post by vix.xvi
is this a gcse question?

A level :smile:
(Original post by flowerscat)Calculate the pH of the solution formed when:
a) 20cm^3 of 0.10moldm-3 HCl is added to 30cm^3 of 0.04moldm-3 NaOH

5. If HCl is in excess - the molar ratio of HCl to H+ is 1:1, so find pH by simply substituting HCl concentration in the pH = -log[H+] equation
6. If NaOH is in excess, use the Kw expression to find H+ and then find pH. OH- conc = NaOH conc (1:1 ratio)

Just to Clarify:

5. If HCl is in excess - the molar ratio of HCl to H+ is 1:1, so find pH by simply substituting (EXCESS) HCl concentration in the pH = -log[H+] equation

6. If NaOH is in excess, use the Kw expression to find H+ and then find pH. OH- conc = (Excess) NaOH conc (1:1 ratio)
Reply 8
Avatar for amnahaa
amnahaa
OP
9
Original post by flowerscat
(Original post by flowerscat)Calculate the pH of the solution formed when:
a) 20cm^3 of 0.10moldm-3 HCl is added to 30cm^3 of 0.04moldm-3 NaOH

5. If HCl is in excess - the molar ratio of HCl to H+ is 1:1, so find pH by simply substituting HCl concentration in the pH = -log[H+] equation
6. If NaOH is in excess, use the Kw expression to find H+ and then find pH. OH- conc = NaOH conc (1:1 ratio)

Just to Clarify:

5. If HCl is in excess - the molar ratio of HCl to H+ is 1:1, so find pH by simply substituting (EXCESS) HCl concentration in the pH = -log[H+] equation

6. If NaOH is in excess, use the Kw expression to find H+ and then find pH. OH- conc = (Excess) NaOH conc (1:1 ratio)

It turned out that HCl was in excess :smile: thank you so much for your help!
Original post by amnahaa
It turned out that HCl was in excess :smile: thank you so much for your help!

Glad to have helped :smile:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you