 Let's go through this problem which is asking us to calculate the pH of this solution at 25 degree Celsius and the solution is 20 ml of 0.5 molar aqueous NH3 and 10 ml 0.5 molar aqueous HCl and it is also given that the base ionization constant for NH3 is 1.8 into 10 to the power negative 5. So if you want to give this a try, you can pause here and attempt it and I'll go ahead with the solution in a few seconds. Okay, so let's start. Now we know that NH3 is a weak base and HCl is a strong acid. So we know that an acid base neutralization will happen. So first let us calculate the number of moles of NH3 and the number of moles of HCl which can take part in the reaction. So to calculate the number of moles, we know that molarity is given in moles per liter. So we can write the number of moles of NH3 divided by the volume of NH3 in liters will be equal to the molarity which is this 0.5. So if we solve this, we get the number of moles of NH3 to be 0.5 times 0.02 which is 0.01 and similarly we can also calculate the number of moles of HCl which we can calculate like this and we find it out to be 0.005. So now that we have the number of moles of NH3 and number of moles of HCl, let's write down the reaction and see how many moles of the reactants are consumed in the reaction. So if we write down the reaction, it will look something like this where we have aqueous NH3 reacting with the H plus from the acid giving NH4 plus. Now instead of writing this as H3O plus, you can also interchangeably light it as H plus but the point is this H plus or H3O plus is what is coming from a strong acid here and which is reacting with the weak base. So now for this reaction, if you look at what were the initial number of moles of the reactants present in the solution, we had 0.01 moles of NH3 and 0.005 moles of the acid that is H plus or H3O plus and these we saw how we calculated before and initially since no amount of product is formed, the initial number of moles of NH4 positive is 0. Now as the reaction happens, if we look at what is the change in the number of moles, we know that since the acid has the lower number of moles, this will be our limiting reagent. What will happen is all of the moles of the acid will be consumed. So the change in H3O plus will be minus 0.005 and because the mole ratio here is 1 is to 1 between the reactants, the same number of moles of NH3 will be consumed. So the change in number of moles of NH3 will also be minus 0.005 and again because the mole ratio here is 1 is to 1, the number of moles of NH4 positive will be plus 0.005. Now the minus and the plus here denotes that in the case of minus, the reactant is being consumed and when I have used a plus, it indicates that a product is formed. So now that we know the initial number of moles and the change, if we look at the final number of moles, we have all of the acid that is getting consumed. So the final moles of H3O plus is 0 and for the ammonia here, we have 0.01 minus 0.005 which is equal to 0.005 and similarly for NH4 positive, the change is plus 0.005. So the final number of moles is 0.005. So this what we have here is the final number of moles of NH3, H3O plus and NH4 positive. So as you can see in the solution, after the reaction, we will have some ammonia and we will have some NH4 positive. So if we were to calculate the molar concentration of NH3 and NH4 positive, we can calculate it as molarity is given by moles per volume. So we know the number of moles. So if you look at NH3, we know it is 0.005 divided by the total volume which is 30 ml, that is this 20 plus 10, which we can write as 0.03 litres. So the molar concentration of NH3 will be 0.167 molar. And similarly for NH4 positive, we can calculate it and because the numbers are the same, the value of the molar concentration of NH4 positive ions will also be equal to 0.167 molar. So now that we have these molar concentrations and we know that there is some amount of NH3 left in the solution, we know that this NH3 will be in equilibrium with the water, that is there in the solution. So let's write down that reaction and see what happens there. So we know that the ammonia in the solution is going to be in equilibrium with water, that is already present in the solution. So I can write this reaction like this, where ammonia is in equilibrium with water, forming hydroxide ions and NH4 positive. And now for this reaction, if we look at the initial molar concentrations, as we just saw, we already have 0.167 molar NH3 in the solution along with NH4 positive, which also has the same molar concentration. And initially let's say the concentration of hydroxide ions is 0. Now as the reaction proceeds and it reaches an equilibrium, if we look at what will be the change in the molar concentration and let's say that the change in the molar concentration is minus x. And because mole ratios of this NH3 and this hydroxide ion is 1 is to 1, we know that the amount of OH negative ions formed will be equal to plus x. And again since the mole ratio between these two is also 1 is to 1, the change in the NH4 plus ions will also be plus x. So now if we calculate the molar concentration at equilibrium for ammonia, it will be 0.167 minus x. For the hydroxide ions, it's going to be plus x. And for the NH4 positive ions, it's going to be 0.167 plus x. So now that we have the molar concentrations at equilibrium, we can write the equilibrium constant as the molar concentration of the hydroxide ion times the molar concentration of NH4 positive divided by the molar concentration of NH3. And we just saw what the molar concentrations are at equilibrium. So if we plug in these values, we get Kb is equal to the molar concentration of the hydroxide ions times 0.167 plus x divided by 0.167 minus x, which we know from here. Now an important point here is that ammonia is a weak base. So its ionization is going to be very low. So the change in molar concentration here is going to be very small compared to the initial value. So x is going to be very small compared to this 0.167. So if x is a very small number, adding x to 0.167 or subtracting x from 0.167 is effectively going to be the same value because it's so small, it's not going to make a difference. So if we neglect this x here, we can just cancel out this 0.167 from the numerator and the denominator and we can write the value of the concentration of the hydroxide ions as equal to the base ionization constant, which is given as 1.8 into 10 to the power minus 5. So now that we know the value of the molar concentration of the hydroxide ions, let's see how we can calculate the pH of the solution. So we just calculated the molar concentration of the hydroxide ions as 1.8 into 10 to the power minus 5 and to calculate the pH, we are going to be using two relations which is first the definition of POH which is given by negative log molar concentration of the hydroxide ions and we also know that the sum of pH and POH will be 14. So from these two relations, we know that the pH will be equal to 14 minus the POH and if we plug in the value of POH as minus log of molar concentration of hydroxide ions, we get the pH to be 14 plus log molar concentration of OH negative and we already know this value. So if we calculate it, this value comes out to be negative 4.744 and so we know that the pH will be equal to 9.256.