Help!Calculate the pH, graph the resultsof amounts of 0.1M NaOH have been aded to 50.00 mL of 0.1 m HCl.?
a) 49.00 b) 49.90 c) 50.10 d) 51.0 mL
does anyone know how to do this?
Answer:
a)moles NaOH = 49.00 x 0.1 / 1000 = 0.0049
moles HCl = 50.00 x 0.1 / 1000 = 0.005
moles H+ in excess = 0.005 - 0.0049 = 0.0001
total volume = 50 + 49 = 99 mL = 0.099 L
[H+] = 0.0001 / 0.099 = 0.0010
pH = - log [H+] = 2.99
b) moles NaOH = 49.9 x 0.1 /1000 = 0.00499
moles HCl = 0.005
moles H+ in excess = 0.005 - 0.00499 = 0.00001
total volume = 99.9 mL = 0.0999 L
[H+] = 0.00001 / 0.0999 = 0.000100 M
pH = 4.0
c ) moles NaOH = 50.10 x 0.1 /1000 = 0.00501
moles HCL = 0.005
moles OH- in excess = 0.00001
total volume = 100.1 mL = 0.1001 L
[OH-] = 0.00001 / 0.1001 = 0.0000999
pOH = - log [OH-] = 4
pH = 14 - pOH = 10
d) moles NaOH = 51.0 x 0.1 /1000 = 0.0051
moles HCl = 0.005
moles OH- in excess = 0.0001
Total volume = 101 mL = 0.101 L
[OH-] = 0.0099 M
pOH = 3.0
pH = 14 - 3 = 11
I can't understand your choices... pH values should range from 1 to 14... but I don't think that's the case... If you are looking though to the volume of sodium hydroxide needed to reach the equivalence point (where the solution is most likely neutral), I'll guess it's more of 50mL is to ~50 mL, granted that one is a strong acid and another is a strong base.
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