A 100.0 mL sample of a solution that is 0.100 M in HCl and 0.100 M in HCN is titrated with 0.100 M NaOH. Calculate the pH after the addition of the following volumes of NaOH: (b) 75.0 mL

Video transcript

Hello everyone. Today. We are being asked the following question. A 200 mL solution contains 2000.1 50 moller of hydrogen iodide 500.100 moller of hipaa Cloris And undergoes titrate in with 0.100 moller potassium hydroxide. What is the ph of the solution? After the addition of 50 millions of potassium hydroxide? So first we want to make note of what we have. We have our hydrogen iodide, which is a strong acid And because it is a strong acid is going to 100% associate, it's going to associate. And then we have our hypo Cloris and hipaa Cloris on the other hand, is a weak acid. So much so in fact that if you look in a reference text you can find its K. A equaling three times 10 to the negative eighth. So since this K. Is very small, it's dissociation doesn't really contribute much to the concentration of the acidity of the hydrogen protons in the solution. So therefore we can say that the concentration of our protons or R H. D. O plus is going to equal the initial concentration of just our hydrogen iodide And that is going to equal 0.150.150 moller. So we're gonna keep that number from this number. We can go ahead and find the moles of our acid and most of our base. So first we'll find the moles of our protons or acid. We're going to take the polarity which is in terms of moles per liter. So it's going to be 0.150 moles per one leader and then we're going to multiply that by the volume which was 200 mila leaders. Of course we have to convert this and get rid of the middle leaders and leaders. So we have to use the conversion factor that one mL equal to 10 to the negative third leaders. Our units will cancel out and we will be left with 0.3 moles of our HBO plus or hydro ni um ions. Next we need to find the moles of our base or her hydroxide. We're gonna take the clarity of our base Which was said to be for potassium hydroxide which was zero 0.100 Most per liter. Once again we're gonna multiply by the volume. However we're gonna multiply by the 50 millions that we're adding of the base. And so we're gonna get rid of the units and multiplied by the conversion factor That we did before. And we're going to get 0.005 moles of hydroxide. So now that we have all of this information, what can we do with it? We can go in and construct our ice table. So we're gonna take our acid R H 30 plus our hydro ni um we are going to react that with our base or our hydroxide ions. This is actually gonna give us two moles of liquid water. So informing our ice table we have our initial change and our equilibrium. The initial concentrations for our protons was 0.03 moles And four, our hydroxide. It was 0.005 moles. Since we are adding the base to the solution, we're gonna go ahead and subtract that from what we have here. So we're going to subtract both of these by 0.05. And of course water doesn't get any in the calculation because it is a constant. And so when we do that, we get 0.025 moles for our hydrogen concentration. Then we get zero moles for our hydroxide solution. And our last step on the right here is going to be to finally calculate the concentration of our high Joni um ions. So how are we going to do that? Well, we're going to take our Number of moles that we have 0.025 moles and divide that by our total volume. So our total volume will be the 200 ml Plus the ml. And then we're gonna convert those middle leaders into leaders by using the conversion factor not one. Middle leader is equal to 10 to the negative third leaders. And that is going to give us our final answer. Or that's going to give us our polarity of 100.1 molar. And then the question asked us to find a ph and we're going to find a ph down here. The ph is going to be equal to the negative log of our hydrogen concentration. We're gonna plug in the negative log of our 0.1 molar. And this is going to give us a final ph of one as our final answer. Overall, I hope this helped, and until next time.