# calculating pH alevel help PLEASEWatch

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#1
Calculate the pH of the solution formed when 100 cm3 of 0.050 mol dm-3 HCl is added to 50 cm3 of 0.500 mol dm-3 KOH.

So I did:
moles of HCl: (100x10-3) x 0.050 = 5x10-3
moles of KOH: (50x10-3) x 0.500 = 0.025

excess = KOH
0.025 - (5x10-3) =0.02
0.02 / (150x10-3) = 0.134
ph = -log(0.134) = 0.88

???? answer is supposed to be 13.12
2
1 month ago
#2
Hi,
Looks like you have mixed up your signs (+/-) OR H+/OH- somewhere cos the maximum pH of 14 minus your answer (0.88) gives the correct answer (13.12).

Any chemistry nerds around??
1
1 month ago
#3
(Original post by medapplicant2020)
Calculate the pH of the solution formed when 100 cm3 of 0.050 mol dm-3 HCl is added to 50 cm3 of 0.500 mol dm-3 KOH.

So I did:
moles of HCl: (100x10-3) x 0.050 = 5x10-3
moles of KOH: (50x10-3) x 0.500 = 0.025

excess = KOH
0.025 - (5x10-3) =0.02
0.02 / (150x10-3) = 0.134
ph = -log(0.134) = 0.88

???? answer is supposed to be 13.12
0.02 are the moles of OH- ions, not H+. Hence, the OH- ion conc in the solution is 0.13

From here you can use Kw to calculate the pH of the solution.
1
1 month ago
#4
(Original post by medapplicant2020)
Calculate the pH of the solution formed when 100 cm3 of 0.050 mol dm-3 HCl is added to 50 cm3 of 0.500 mol dm-3 KOH.

So I did:
moles of HCl: (100x10-3) x 0.050 = 5x10-3
moles of KOH: (50x10-3) x 0.500 = 0.025

excess = KOH
0.025 - (5x10-3) =0.02
0.02 / (150x10-3) = 0.134
ph = -log(0.134) = 0.88

???? answer is supposed to be 13.12
I would say it is correct, worked it out and got the same answer.
1
1 month ago
#5
0.13 is [OH¯] now you need to use Kw (10^-14) and [OH¯] to find [H+] and then use pH formula
(Original post by medapplicant2020)
Calculate the pH of the solution formed when 100 cm3 of 0.050 mol dm-3 HCl is added to 50 cm3 of 0.500 mol dm-3 KOH.

So I did:
moles of HCl: (100x10-3) x 0.050 = 5x10-3
moles of KOH: (50x10-3) x 0.500 = 0.025

excess = KOH
0.025 - (5x10-3) =0.02
0.02 / (150x10-3) = 0.134
ph = -log(0.134) = 0.88

???? answer is supposed to be 13.12
1
#6
(Original post by Cant do it)
0.13 is [OH¯] now you need to use Kw (10^-14) and [OH¯] to find [H+] and then use pH formula
Thanks everyone that really helped, now I'm stuck on this question lmao

Calculate the pH of the solution formed when 50cm3 of 0.5moldm3 propanoic acid (pka = 4.87) is added to 100cm3 of 0.080 moldm3 of KOH

so I worked out the moles of the acid : 0.025 and the base: 0.008
acid is in excess so: 0.025 - 0.008 = 0.017
concentration of acid: 0.017/150x10-3 =0.113
then to work out H+ : the square root of (4.87x10-3) x 0.113 = 0.023
pH = -log10(0.023) = 1.64
UGHHHH answer is supposed to be 4.54
0
1 month ago
#7
(Original post by medapplicant2020)
Thanks everyone that really helped, now I'm stuck on this question lmao

Calculate the pH of the solution formed when 50cm3 of 0.5moldm3 propanoic acid (pka = 4.87) is added to 100cm3 of 0.080 moldm3 of KOH

so I worked out the moles of the acid : 0.025 and the base: 0.008
acid is in excess so: 0.025 - 0.008 = 0.017
concentration of acid: 0.017/150x10-3 =0.113
then to work out H+ : the square root of (4.87x10-3) x 0.113 = 0.023
pH = -log10(0.023) = 1.64
UGHHHH answer is supposed to be 4.54
Propanoic acid is a weak acid so you'd have to use ka formula. do you know how to do that?
0
#8
(Original post by Cant do it)
Propanoic acid is a weak acid so you'd have to use ka formula. do you know how to do that?
Yeah so Im using ka = (H+)2 / HA

rearranged so its the square root of KA x HA = H
so i got 1.35x10-5 x 0.113
but im not getting the right answer having a serious breakdown
0
1 month ago
#9
(Original post by medapplicant2020)
Yeah so Im using ka = (H+)2 / HA

rearranged so its the square root of KA x HA = H
so i got 1.35x10-5 x 0.113
but im not getting the right answer having a serious breakdown
You can only use that formula if the acid has not yet been neutralised but in the question it has so you need to use ka=([propanoate]*[H+]) /[propanoic acid]
1
#10
(Original post by cant do it)
you can only use that formula if the acid has not yet been neutralised but in the question it has so you need to use ka=([propanoate]*[h+]) /[propanoic acid]
omggggg i have been trying to do this question for like an hour now and you have just saved me thanks x
0
1 month ago
#11
(Original post by medapplicant2020)
Yeah so Im using ka = (H+)2 / HA

rearranged so its the square root of KA x HA = H
so i got 1.35x10-5 x 0.113
but im not getting the right answer having a serious breakdown
0
1 month ago
#12
(Original post by medapplicant2020)
omggggg i have been trying to do this question for like an hour now and you have just saved me thanks x
Lmao np happy to help anytime
0
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