I reallllllly need help with my post-lab questions! It would be great if I could
ID: 812708 • Letter: I
Question
I reallllllly need help with my post-lab questions! It would be great if I could get an explanation for all of these but I atleast would like 3 worked out and explained for Part I and Part II so I can get the gist of it and do it on my own. Thank you in advance!
50 mL of 0.20M NaC2H302, 10 mL of 1.0M HC2H3O2, and 40 mL of H2O were mixed to create a buffer with a pH of 4.74 (Ka= 1.8 x10^-5), demonstrated by the equation:
HC2H3O2 (aq) + H2O (l) <====> (H3O)+ (aq) + (C2H3O2)- (aq)
Part. I:
a.) What is the pH when 5 mL of 0.10M HCl is added to the buffer?
b.) When 10 mL of 0.10M HCl is added to the buffer?
c.) When 20 ml?
d.) When 30 mL?
e.) When 50 mL?
f.) When 100 ml?
Part II:
a.) What is the pH when 5 mL of 0.10M NaOH is added to the buffer?
b.) When 10 mL?
c.) When 20 mL?
d.) When 30 mL?
e.) When 50 mL?
f.) When 100 mL?
QUESTION: Part III asks to calculate the same additions of HCl and NaOH but instead of adding to a buffer, I have to add it to water. How would I approach solving the pH for this?
Explanation / Answer
pH of buffer = 4.74
[H+] in the buffer = 10-4.74
= 1.82 * 10-5 moles
when 5 mL of 0.10M HCl is added to the buffer, moles of HCl added = mole sof protons as HCl is a strong acid
moles of HCl = [H+] = 0.10 moles/litre * 0.005 litre = 0.0005 moles
so these moles of H+ ions add to the moles of protons already present in the buffer
So total [H+] now = (0.0000182 moles) + 0.0005 moles
= 5.18 * 10-4 moles
so pH = - log (5.18 * 10-4 )
= 3.28
Similarly calculate moles of HCl for each addition, add those moles to already moles of H+ ions in buffer and then find pH of the total H+ ions
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Part II:
moles of NaOH = 0.10 moles/litre * 0.005 litre = 0.0005 moles
these moles of NaOH neutralize same moles of H+ ions present in the buffer
So the remaining moles of H+ ions in the buffer = (0.0000182 moles) - 0.0005 moles
= 4.818 * 10-4 moles
Its these moles of H+s remaining un-neutralied contribute to pH
so pH of the bufer after adition of these 5 mL of 0.10M NaOH are added = - log (4.818 * 10-4 )
= 3.32
Similarly calculate moles of NaOH added in ecah case, find equivalent moles of H+ ions neutralized, subtract these moles from total to get remianing moles of H+s which are in buffer and finally calculate pH