a) Calculate how many cm3 of 1.00 moldm-3 HCL are needed to decrease the pH of 1.00 dm3 of HCL by one unit. Ignore the small change in volume
b) Calculate how many cm3 1.00moldm-3 NaOH are needed to increase the pH of 1.00dm3 of HCL by one unit. Ignore the small change in volume
2 answers
b. concentration of 1.00 mol dm-3 NaOH = 1mmol cm-3 ; so 0.9 cm3 has 0.9mmol.
Don't you have a starting pH of the 1 dm3? At a pH of anything smaller than pH = 3 the volume added can't be ignored. If you assume pH = 3 and you want it to be 1 unit less or pH = 2 then you want the make the final concn 10 times greater. I would look at it this way.
You have pH = 3 or (H^+) = (HCl) = 0.001 so you have 1000 mL x 0.001 = 1 millimole HCl. You want 10 millimoles (since 10 mmol/1000 mL - 0.01) so you will need to add 9 millimoles to do that. You can get that from
9 mL of 1 M HCl (that's 1 mol/dm3
The NaOH is done the same way. Post your work if you get stuck.
You have pH = 3 or (H^+) = (HCl) = 0.001 so you have 1000 mL x 0.001 = 1 millimole HCl. You want 10 millimoles (since 10 mmol/1000 mL - 0.01) so you will need to add 9 millimoles to do that. You can get that from
9 mL of 1 M HCl (that's 1 mol/dm3
The NaOH is done the same way. Post your work if you get stuck.
0 answers