A 1.0-L buffer solution is 0.125 M in HNO2 and 0.145 M in NaNO2. Determine the concentrations of HNO2 and NaNO2 after the addition of each substance: a. 1.5 g HCl

Video transcript

Hey, everyone. And welcome back to another video. A 1 L buffer solution is 0.125 molar in nitrous acid and 0.145 molar in sodium nitrate determine the concentrations of nitrous acid and sodium nitride. After the addition of each substance, a 1.5 g of HCL and we are given four answer choices. ABC MD, those sensors have different concentrations of HNO two and NANO two. So we're going to solve this problem. Let's do that. First of all, we understand that we have a buffer solution and we're adding a strong acid. Let's remember that hydrochloric acid HCL is a strong acid. So we can use a shorthand notation because we have an equivalent amount of hydro num. And because we have a buffer solution, we need to choose the basic component which reacts with a strong acid. So we know that the base is nitrite and ion, right? Because sodium dissociates, it's an spectator in a solution. So now I tried reacts with hydro to produce nitrous acid, right? We can write HNO two aqueous. And what we can do here is just introduce an IC table which will be helpful for us in determining the initial change and equilibrium concentrations for this problem. Now what about the initial concentrations? We have 0.125 molar for nitrous acid, we also have 0.145 molar of sodium nitrate, which is equivalent to nitride as well. Right, we have a 1 to 1 ST geometry and now we want to identify the molarity of hydro which is equivalent to the molarity of HCL. So first of all, we need to take the number of modes of HCL and divide that by volume. Now we have 1.5 g, we need to divide that by the molar mass of HCL. That would be 36.46 g per mole. And we can divide everything by the volume which is 1 L, right? Because we get moles, we divide it by volume, we get molarity and we end up with 0.041 one molar. So this is the molarity of hydro 0.0411 molar. And we can clearly see that this is the limiting reactant because we have a 1 to 1 reaction. So the lower amount represents the limiting reactant, meaning it reacts fully. And this is our change. And therefore, at the end, we have zero molar of H plus, right is the limiting reactant and we preserve our buffer. And according to ST geometry, we're going to use negative 0.0411 molar or nitrate and plus 0.0411 molar for the acid, right? Because it's the product. So we are using a positive sign and now we just want to calculate the final amounts. So starting with no two negative, we can use three decimal places, right? Three significant figures here. So we would have an equilibrium molarity of zero point 104. And for the acid, we end up with 0.166 molar and that's it. Right. We got our number. So let's see which one is the correct answer. We noticed that if the concentration of nitride is 0.104 then the correct answer should be d because it's the only answer that has this concentration of sodium nitrate. The correct answer is option D 0.166 molar of H and 02 and 0.104 molar of nano two. That's our final answer. And thank you for watching.