1. Determine the pH during the titration of 13.4 mL of 0.114 M hydrobromic acid
ID: 706041 • Letter: 1
Question
1. Determine the pH during the titration of 13.4 mL of 0.114 M hydrobromic acid by 4.00×10^-2 M sodium hydroxide at the following points:
(a) Before the addition of any sodium hydroxide
(b) After the addition of 19.1 mL of sodium hydroxide
(c) At the equivalence point
(d) After adding 45.8 mL of sodium hydroxide
2. Determine the pH during the titration of 17.3 mL of 0.293 M nitric acid by 0.182 M sodium hydroxide at the following points:
(a) Before the addition of any sodium hydroxide
(b) After the addition of 13.9 mL of sodium hydroxide
(c) At the equivalence point
(d) After adding 35.9 mL of sodium hydroxide
PLEASE ANSWER BOTH QUESTIONS!
Explanation / Answer
HBr = 13.4ml of 0.114M
number of moles of HBr = 0.114Mx0.0134L=0.0015276 moles
a) Before the addition of NaOH
HBr is a strong acid. So the Concentration of H+ is equal to Acid
[H+] = 0.114M
-log[H+]= -log[0.114]
PH= 0.943
b) afterthe addition of 19.1mL of NaOH
Concentration of NaOH= 4.00x10^-2M=0.04M
number of moles of NaOH= 0.04Mx0.0191L= 0.000764 moles
number of moles of Acid is greaterthan the Base
SO the resultant concentration of solution is equal to H+
Remaining number of moles H+ = 0.0015276 - 0.000764=0.0007636 moles
Total volume= 13.4+ 19.1=32.5 ml = 0.0325 L
Concentration of H+ = number of moles/volume in L = 0.0007636/0.0325 = 0.0235 M
[H+] = 0.0235M
-log[H+] = -log( 0.0235)
PH=1.628 = 1.63
PH= 1.63
C) at equivilent point PH= 7
d)after addition of 45.8 ml of NaOH
number o f moles of NaOH = 0.04Mx0.0458L= 0.001832 moles
number of moles of Base is greaterthan acis.
So the concentration of resultant solution is equal to OH-
remaining number of moles of OH- = 0.001832 - 0.0015276=0.0003044 moles
Total volume = 13.4+45.8 = 59.2ml = 0.0592L
[OH-] = 0.0003044/0.0592=0.00514M
[OH-] = 0.00514M
-log[OH-] = -log( 0.00514)
POH= 2.289
PH+POH= 14
PH= 14-POH
PH= 14- 2.289
PH=11.711
PH= 11.7
The second problem is also similar manner.
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