ATAR Notes: Forum
HSC Stuff => HSC Science Stuff => HSC Subjects + Help => HSC Chemistry => Topic started by: preliminary17hsc18 on July 28, 2018, 03:36:27 pm
-
Hi all,
I just had a question I need some help with:
Calculate the pH after 20mL of 0.01M NaOH is added to 50mL of 0.2M HCl
Also, what pH would the resulting solution be if 10mL of 0.005M sulfuric acid is diluted to 100mL?
Thank youuu!
-
Question 1
NaOH + HCl -----> NaCl + H2O
Moles of NaOH = VOLUME TIMES CONCENTRATION; 0.02L times 0.01 = 0.0002 moles
Moles of HCL = 0.05L times 0.2 = 0.01 moles
Therefore, HCl is the excess reagent whereas NaOH is the limiting reagent
Because of this, only 0.0002 moles of HCl is used up in the reaction
So that means the moles of HCl left would be 0.01-0.0002 which equals 0.0098 moles
The resultant solution has a volume of 70 mL or 0.07 litres
Therefore, the concentration of the resultant solution would be 0.0098 moles divided by 0.07 litres
This equals 0.14
And to calculate pH we use the logarithmic formula; -log(0.14) which approximately equals 0.85
Question 2
You simply use this formula
C1V1 = C2V2
0.01 Litres times 0.005 = 0.1 times C2 (concentration we dont know)
C2 = (0.01 times 0.005) divided by 0.1; C2 = 0.0005
H2SO4 is a diprotic acid, meaning it releases up to 2 hydrogen ions per molecule
Therefore, when calculating its pH we simply have to times the 0.0005 by 2
Which makes -log(2 times 0.0005); -log(0.001)
pH = 3