1pH of solution X is 4 -> ( pKa1 + pKa2 )/2
This may be simple but I'm not too comfortable with ionic(even the simplest[2][2]). So pls help !!!
A young scientist performs the foll. set of experiments to find the pH of the resulting soln. at diff. equivalence pts. ,the vol. of reagent used & the total vol. of the soln.
He titrates 100ml of 1M H4A ( Ka1 = 10-3 , Ka2 = 10-5 , Ka3 = 10-9 , Ka4 = 10-13 ) with M/2 NaOH soln.
The various stages of the exp. are :
At 1st equivalence pt.,he forms soln "X" by addition of M/2 NaOH into 100ml 1M H4A soln. , then he forms soln. "Y" by addition of same NaOH in "X" at 2nd equivalence pt.Similarly , soln. "Z" is formed by addition of same NaOH in "Y" at 3rd equivalence pt.Last,he forms soln. "W" by addition of same NaOH in "Z" and additional 100ml H2O is added to "W".
1) What is the pH of soln. "X" ?
A] 7 B] 3 C] 5 D] 4
2) What is the vol.(in ml) of soln. "Y" ?
A] 700 B] 500 C] 100 D] 300
3) What is the rate const.(in L mol -1 s -1) of the forward reaction of soln. "W" if the rate const. of the backward reaction is 3 x 102 L mol -1 s -1
A] 30 B] 3 x 105 C] 0.3 D] 3 x 102
Pls. also post solns.(not X,Y,Z & W [4][4]) along with answers.
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2 Answers
1Vol. of Y is 500 ml -> 100 + 200 (NaOH added first time) + 200 (NaOH added 2nd time )