Here it says to calculate the pH of the solution resulting from the titration between 25 MLS of a .100 molar chloric acid solution and 50 mills of a .100 molar ammonia solution. Here we're told that KB value of ammonia is 1.75×10-5. All right, so we're going to use the steps one to three to help set up the ICF chart. Now, if you don't remember the steps of one to three, make sure you go back and take a look at my video dealing with the titration between a weak acid and a strong base.

We're going to say here that we have a weak species and a strong species, so we have to set up an ICF chart. We're going to say that the strong species has to be set as a reactant. So here chloric acid has to be reacted. It reacts with its chemical opposite. What's the opposite of an acid? A base. The ammonia is the base. Now using the Bronsted-lowry definition of acid and bases, the acid will donate an H plus to the base. So it donates 1 to ammonia to give us ammonium ion. And then what we have left here is the chlorate ion.

Now in an ICF chart we only care about three things. We only care about the weak acid, or in this case conjugate ask the same thing. We only care about the conjugate base or weak base and whatever is strong. The fourth thing we ignore it now in an ICF chart, the units have to be in molds, which is liters times molarity. So divide the MLS by 1000, multiply by the molarity. So that will give us 0025 moles of our strong acid, 0050 moles of our weak base 0 initially of this conjugate acid slash weak acid.

Look at the reactant sign. The smaller moles will subtract from the larger moles, so 0.0025 or minus from both. We know we're dealing with an ICF which is initial change final, so at the end we'll have .00250 left of this. Remember, based on the law of conservation of mass, matter is either created nor destroyed. It just changes form, which means whatever we lose on the reactant side, we're gaining on the product side. So we're gaining 0.0025 here.

So we're going to say here at the end what do we have? We have a weak acid and conjugate based remaining, so that leads us to Step 4. The Henderson Hasselbeck equation is used for a buffer to find the pitch of a solution. And that's what we have. We have a buffer. Now using the final row, use the moles of the weak acid and its conjugate base to find a pH. Now the Henderson houseboat can be observed two different ways where it could be pH equals PKA plus log of conjugate based over weak acid. Or to simplify things for us, we can use this other equation when given KB or pH equals PKB plus log of conjugate acid over weak base.

So since we're giving KB in the question, let's use the second version. So pH equals PKA on PKB plus log of conjugate acid over week base. So we're going to say here equals negative log of 1.75×10-5 plus log of point 0025.0025. When you plug that in, you're going to get 4.76 as the PH-4 solution. That's before the equivalence point.