Calculate the fraction of the intermediate for sulfurous acid, H 2 SO 3 , at pH = 11.00? pK a1 = 1.80, pK a2 = 7.19.

Here we have to calculate the fraction of the intermediate for sulfuric acid at pH of 11.00. Now we're told that the pKa1 value is 1.80 and the pKa2 value is 7 point 19. Now, remember for a sulfuric acid, it's a diprotic acid. We have the acid form, we have the intermediate form and we have our basic form. Here, pH would equal pKa 1. PH here would equal pKa 2. We can see here that the pH is 11.0 which would put it beyond this line. The major form would exist as the basic form. Here, we're just looking for the fraction of the intermediate form though, which would be our second greatest amount. Using the formula that we know for intermediate form, we'd say the fraction of the intermediate form equals k a one times h+ivided by h+quared+h+timeskaone pluska1 timeska2. So remember, h plus is equal to 10 to the negative PH. So that'd be 10 to the negative 11. We'd say here that ka1 equals 10 to the negative pka1 one which is 10 to the negative 1.80 and k a 2 is equal to 10 to the negative p k a 2 which is 10 to the negative 7.19. Here, we input those values. So k a one, so 10 to the negative 1.80 times 10 to the negative 11.00 divided by 10 to the negative 11.00 squared plus 10 to the negative 11 times 10 to the negative 1.80 plus 10 to the negative 1.80 times 10 to the negative 7.19. So when we solve for what's on the top and the bottom, we're gonna get 1.58 489 times 10 to the negative 13 divided by 1.02345 times 10 to the negative 9. When we punch that in, we get 0.000155 as the fraction for the intermediate form of this solution at a pH of 11.00. It's an incredibly small number which is expected because at a pH of 7.19 which is equal to pKa 2, we'd have 50% of the intermediate form and the basic form. Since the pH is several units above that, we're going to have a huge amount of the basic form and very little of the intermediate form and even smaller amount of the acid form. Remember, the pH of the solution has a direct impact on the amount of each one of these forms for diaprotic, polyprotic or monoprotic species. Utilizing the formulas will just help you figure out what's the exact value for each one of the species found in your given solution.