Calculate the pH of the solutions resulting from the following reactions.
25cm³ of 0.1M HCL reacting with 5cm³ of 0.15M NaOH
For the [H+] i got 121 and for OH− i got 0.025 final pH i got was 1.03 to 2dp
Where did it all go wrong?
You've forgotten to work out the moles of excess acid. Firstly, you work out the moles of acid and alkali. ACID = 2.5x10^-3 ALKALI =7.5x10^-4 But the acid is in excess, so you need to find how many unreacted moles are in the resulting solution. EXCESS ACID= ACID-ALKALI =1.75x10^-3
Now, work out the concentrations using the final volume, 30cm^3. ACID= 0.583
You've forgotten to work out the moles of excess acid. Firstly, you work out the moles of acid and alkali. ACID = 2.5x10^-3 ALKALI =7.5x10^-4 But the acid is in excess, so you need to find how many unreacted moles are in the resulting solution. EXCESS ACID= ACID-ALKALI =1.75x10^-3
Now, work out the concentrations using the final volume, 30cm^3. ACID= 0.583
So i never needed to work out hydrogen ion concentration??
How do i know acid is in excess and not the base?
Because the acid will have more moles. Sometimes the base is in excess and Kw=[OH-][H+] will be needed, where you rearrange for H+. In this case, the excess acid was a strong one so the [HA] is taken as the [H+], but if it was H2SO4, or any other diprotic acid, the moles would need to be doubled. If you had an excess of a weak acid, then Ka would come in and that's when you rearrange for H+.
Because the acid will have more moles. Sometimes the base is in excess and Kw=[OH-][H+] will be needed, where you rearrange for H+. In this case, the excess acid was a strong one so the [HA] is taken as the [H+], but if it was H2SO4, or any other diprotic acid, the moles would need to be doubled. If you had an excess of a weak acid, then Ka would come in and that's when you rearrange for H+.
hmmmm, i don't understand >.> part of this is the teachers fault for not teaching me this >.>
hmmmm, i don't understand >.> part of this is the teachers fault for not teaching me this >.>
It all depends on the situtation that then determines the calculation to use. pH calcs are so repetitive, so practice themand you'll see a pattern emerge :-) my teacher was poo too.
It all depends on the situtation that then determines the calculation to use. pH calcs are so repetitive, so practice themand you'll see a pattern emerge :-) my teacher was poo too.
I'm gonna go with more moles of it calculated if that's not it i'll have to ask another chem teacher, shame really because i got the worst out of the bunch
I'm gonna go with more moles of it calculated if that's not it i'll have to ask another chem teacher, shame really because i got the worst out of the bunch
This is not rocket science.
1. Calculate the moles of both the acid and the base. 2. The one with the larger number is in excess. 3. Subtract the smaller number from the larger number of moles and you have the excess moles. 4. If the acid is in excess, divide the moles excess acid by the volume to get the molarity. Use this to calculate pH 5. If the base is in excess, divide the moles excess base by the total volume to get the molarity of the base. Calculate pOH (same method as pH except using [OH-] instead of [H+], The subtract result from 14 to get pH.
NOTE: This is only this simple when the balanced equation is 1:1
1. Calculate the moles of both the acid and the base. 2. The one with the larger number is in excess. 3. Subtract the smaller number from the larger number of moles and you have the excess moles. 4. If the acid is in excess, divide the moles excess acid by the volume to get the molarity. Use this to calculate pH 5. If the base is in excess, divide the moles excess base by the total volume to get the molarity of the base. Calculate pOH (same method as pH except using [OH-] instead of [H+], The subtract result from 14 to get pH.
NOTE: This is only this simple when the balanced equation is 1:1
So if it's a 2:1 in favour of H+ ions then i multiply the value i get from step 4 by 2?
No, you must factor in the 2:1 ratio when calculating the excess.
For example if you react 25ml of 0.1M sulfuric acid with 10ml of 0.2M NaOH,
Mol H2SO4 = 0.0025 Mol NaOH = 0.002
BUT
2 mol of NaOH react with each mol of H2SO4 .
SO
0.002 mol of NaOH reacts with 0.001 mol of H2SO4
therefore the H2SO4 is in excess by 0.0025 - 0.001 = 0.0015 mol
NOW
To find mol of H+ ions you double the sulfuric acid concentration value.
Do i have to half the NaOH mol? or can i double the sulfuric acid mol? then divide that answer by 2?
Oh i see, so i have to adjust the mol when calculating excess too? Ok thanks! Didn't know that, i'll be sure to tag you when i get stuck on some chem ^-^