Here we're told that sulfuric acid, which is H_{2}SO_{3}, represents a diprotic acid with a K_{a1} equal to 1.6 × 10^{-2}. K_{a2} itself equals 4.6 × 10^{-5}. Now it says, calculate the pH and concentrations of sulfuric acid, hydrogen sulfide also known as bisulfite, and sulfate ion when given 0.200 molar of sulfuric acid. For sulfuric acid, we're dealing with the acidic form or the fully protonated form of the compound. And remember, when we're dealing with the acidic form of the compound, we use K_{a1} to help us determine what the pH will be. Now, our diprotic acid is H_{2}SO_{3}. It will react with water. Now remember, the acid is going to donate an H^{+} to the water. It's going to become HSO_{3}^{-} Remember, charged species are aqueous when in solution and water itself becomes H_{3}O^{+}. We have initial change equilibrium to represent the ICE chart that we've set up. Remember, in an ICE chart, we ignore solids and liquids. Water, which is a liquid, will be ignored. Now initially, we have 0.200 molar of my diprotic acid. The products initially are 0. We lose reactants to make products. Bring down everything. At this point again, we're dealing with removing the first acidic hydrogen from our compound. So, we're dealing with K_{a1}. K_{a1} equals products over reactants. So, it's HSO_{3}^{-} times H_{3}O^{+} divided by H_{2}SO_{3}. We're gonna plug in the values that we know. We have 1.6 × 10^{-2} equals. At equilibrium, both my products are equal to x and since they're multiplying each other, that's x^{2} divided by 0.200 minus x. At this point, we have to determine can we ignore the minus x or not. To do that, we do the 5% approximation method. Now, with this 5% approximation method, we take the initial concentration of our diprotic species here and we divide it by the k value that we're using. In this case, K_{a1}. If the ratio is greater than 500, then I can ignore that minus x and avoid the quadratic formula. So, our initial concentration is 0.200 molar divided by our K_{a1} which is 1.6 × 10^{-2}. When we do that, we get back 12.5. We have a value here that is not greater than 500, so we cannot ignore the minus x, and we're gonna have to do the quadratic formula. So, we're going to multiply both sides here by 0.200 minus x. Here, we're going to distribute, distribute. Here, that's going to give me 0.0032 minus 1.6 × 10^{-2} x equals x^{2}. Here, this x^{2} has the highest power, so it's going to be our lead term which means everything gets moved over to its side. So, we're going to add 1.6 × 10^{-2} x to both sides and subtract 0 0032 from both sides. When we do that, we're gonna get the expression x^{2} x squared plus 1.6 × 10^{-2} x minus 0.0032. So, this is my a, my b, and my c. My quadratic formula will be -b plus or minus square root of b^{2} - 4ac over 2a. When we plug in the values, so -1.6 × 10^{-2}, plus or minus 1.6 × 10^{-2} squared minus 4 times 1. Don't forget the negative sign of c. So times -0.0032 divided by 2 times 1. So here, when we solve for everything in here and then take the square root of that, what we'll get at this point is x equals -1.6 × 10^{-2} plus or minus 0.114263 divided by 2. Realize here that we have plus or minus which means we're gonna get 2 possible x variable answers. So x equals 0.049131 if we added or x equals -0.065132 if we use the minus instead. How do we know which one to use? Well, at equilibrium, you cannot have a negative concentration and it's not possible. If we use that negative x and plugged it into here or into here for equilibrium, that gives us a negative value. That would mean that that is not the correct concentration to use. This one does not work. My x will be this number here. All we say now at this point is at equilibrium, H_{2}SO_{3} equals 0.200 x. Plug in what we just found in for x and what we get for equilibrium amount for sulfuric acid at equilibrium is 0.151 molar. At equilibrium, hydrogen sulfide or bisulfide equals H_{3}O^{+} because they both equal x which again we said is equal to 0.049131 molar. And because we know what H_{3}O^{+} is, we know what pH is because pH equals negative log of H_{3}O^{+}. So here, that gives me 1.31 as my pH once I plug that in. Now, up to this point, what have we determined? Well, we figured out what the pH is of my diprotic acid. We determined what the final equilibrium amount of sulfurous acid is and at this point, hyd...