Does the pH increase, decrease, or remain the same when the substances are added to the solutions? (a) LiF to an HF solution (b) KI to an HI solution (c) NH4Cl to an NH3 solution

Video transcript

Welcome back everyone. We need to determine whether the Ph will change or not. In the following scenarios, if the Ph changes identify whether it will increase or decrease. So beginning with scenario, a we have sodium bromide added to hydro Tomic acid solution. We should recognize that hydroponic acid is one of our strong assets and being a strong asset, it's going to dissociate into its following ions where we form one mole of proton And one mole of our bromide. An ion. Now focusing on our salt sodium bromide because it's a salt, it's going to dissolve in water and will form the following catalon sodium Catalan and R B R minus an ion. So we need to determine whether our salt will have a effect on the Ph of our solution. And we need to do so by analyzing the ions. Now, first focusing on our sodium catalon, we should recognize that sodium as an element on the periodic table is an alkaline metal and therefore, its ion has no effect on Ph. Now looking at R B R minus an eye on notice that this an ion is also formed when our strong acid hydra Bronek acid releases a proton. And we want to recall the role that the anti informed when a strong acid releases a proton, the an ion. And so we'll just say the anti informed when a strong acid releases a proton is not basic or acidic to any significant extent. And so because of that, we would say that therefore, sodium bromide is going to be overall a neutral salt because it has an influence on Ph equal to seven. And so when our salt, sodium bromide is added to hydroponic acid in solution, the ph will not change and sorry, this says PH will not change. So this would be our first answer for scenario A. Now let's consider scenario B we have Benzel Ammonium chloride C six H five and H three cl added to an Allen C six H five and H two. Now focusing on our aniline compound first C6 H five and H two recall that this is actually considered a weak base. And so it's going to be in when it dissolves in water, it's going to be an equilibrium with its eye on products where in this case, water is acting as our acid donating a proton to an Allen. And we would form the conjugate acid C six H five N H three plus. And here are our labels as well as our second products being hydroxide as the conjugate base of water. So let's note that Arbenz L ammonium carry on is the conjugate acid of our weak base aniline. And more specifically, we can say it's a conjugate, weak acid of aniline, our weak base. And therefore, we can understand that this cat Ian is actually going to influence our ph in an acidic way. But let's move on to the salt, which is our Benzel ammonium chloride C six H five and H three C O, which is going to dissolve in water to form our benzel ammonium catalon as well. And our chloride minus one minus an ion. So note that we have benzel ammonium carry on in both of our reactions here. And so this caddy on is going to be a common ion since it's present in both reactions. And specifically this common ion, Benzel ammonium cat ion is present on the product side of both of our reactions. This means that due to this presence of this acidic common ion, the equilibrium is going to shift in the direction towards the left of our reaction. So towards the products. And so equilibrium shifts to the left. Therefore, our concentration of our products being our benzel ammonium cat ion concentration and our concentration of hydroxide will both decrease. And so because our concentration of hydroxide decreases, and we know our benzel ammonium cologne has an acidic effect on the Ph, we will say that therefore, the PH will also decrease. And so we have a more acidic solution. So this would be what's highlighted in yellow here will be our second answer for scenario B. When we have benzel ammonium chloride added to aniline in solution, the Ph will decrease. Moving on to scenario C we have sodium cyanide added to hydrogen cyanide acid beginning with H C N hydrogen cyanide. We should recognize that this is not on our memorized list of strong acids and is therefore a weak acid. And so when reacting with water, it's going to donate a proton to water and will be in equilibrium with its eye on products where we form the conjugate base of our hydrogen cyanide, weak acid C N minus and hydro ni um as the conjugate base or sorry, the conjugate acid of water. And just to be more detailed, we can specifically label our cyanide an ion product as the conjugate weak base of our hydrogen cyanide, weak acid. And so now let's focus on our salt, sodium cyanide which dissolves in water and forms its ions and A plus and C N minus. Yet again, we have C N minus as our common ion. And we should recognize that because cyanide C N minus is the conjugate weak base of our weak acid, hydrogen cyanide. It's therefore going to influence our solution in a basic way due to this fact that we have common ions being C and minus on both the product side of these reactions just as before equilibrium is going to shift in the direction towards our reactant. So towards the left, meaning that our concentration of our products CN minus and hydro ni um will both decrease when equilibrium shifts left. And so, therefore, because our concentration of our products are decreasing, we know that hydro ni um is particularly strong acid and because it's strongly acidic and we're lowering the concentration of hydro ni. Um And we know that cyanide being the conjugate weak base of our weak acid H C N is going to have a basic influence on our solution. We would say PH will increase. And so it will be more basic. And so this would be our third final answer for scenario C so all of our answers that we've outlined correspond to choice a in the multiple choice. I hope that everything I went through was clear. If you have any questions, please leave them down below. And I'll see everyone in the next practice video.