Asked by Owen
                What	is	the	pH	of	a	0.16M KHSO3 solution	at	298	K?
a. 4.41
b. 1.31
c. 7.00
d. 1.82
e. 3.89
I did the calculations and my answer came out to be 3.89 but apparently the answer is 4.41. Can anybody explain why ? Thnx in advance
            
        a. 4.41
b. 1.31
c. 7.00
d. 1.82
e. 3.89
I did the calculations and my answer came out to be 3.89 but apparently the answer is 4.41. Can anybody explain why ? Thnx in advance
Answers
                    Answered by
            DrBob222
            
    I suggest you repost the problem at the top of the board and this time INCLUDE the k2 for H2SO3. We can't try to obtain a certain answer if we don't know what numbers you are using. 
    
                    Answered by
            Bonnie
            
    Hi,
I assume you used the ICE chart to find the pH of KHSO3, and that's how you ended up with pH of 3.89 as your answer.
Remember that KHSO3 is a polyprotic acid and that KHSO3, specifically, is amphoteric, meaning that it can act as an acid or base. In the question, it does not specify whether KHSO3 will act as an acid or base so you cannot find the pH using the ICE chart.
Instead, you solve this question using the equation:
pH = 1/2 * (pKa1 + pKa2)
Substitute the Ka1 and Ka2 values and solve for the pH:
pH = (1/2) * ( -log(1.5 x 10^-2) + -log(1.0 x 10^-7) ) = 4.41
    
I assume you used the ICE chart to find the pH of KHSO3, and that's how you ended up with pH of 3.89 as your answer.
Remember that KHSO3 is a polyprotic acid and that KHSO3, specifically, is amphoteric, meaning that it can act as an acid or base. In the question, it does not specify whether KHSO3 will act as an acid or base so you cannot find the pH using the ICE chart.
Instead, you solve this question using the equation:
pH = 1/2 * (pKa1 + pKa2)
Substitute the Ka1 and Ka2 values and solve for the pH:
pH = (1/2) * ( -log(1.5 x 10^-2) + -log(1.0 x 10^-7) ) = 4.41
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