Calculate the pH of solutions prepared by:

(d) Mixing equal volumes of 0.20 M HCl and 0.50 M HNO3.

(Assume that volumes are additive.)

Video transcript

Welcome back everyone to another video, a solution of equal volumes of 0.50 molar hydrochloric acid and 0.30 molar nitric acid was prepared. Find its Ph note assume that volumes are additive. First of all, what do we know about these two assets? Well, essentially they are strong assets. So what we're going to do is just understand HCL and HNO three ionize, they ionize completely. So we're just going to add the water and we're going to illustrate the formation of hydro num and the conjugate base for HCL data BC L minus chloride. This process takes place to an extent of 100% because it is a strong acid. And we would essentially have the same scenario for HNO three. It's one of the seven strong acids that we know whenever it ionizes, it releases the same quantity of H 30 plus and nitrate. If we want to identify Ph I work the total concentration of hydro, right. So what we're going to do here is understand that we are taking 0.50 molar of HCL. According to ST geometry, we would also have 0.50 molar of H we O plus formed right from 0.50 molar hcl. For HNO three, we are starting with 0.30 molar. And based on the complete ionization, we would have 0.30 molar of hydro. Now, if we are mixing them, we want to find the final concentration of hydro. How do we do that? Well, essentially we have to recall the formula that we use for concentration can be expressed as the number of moles and divided by volume V. How do we find the total number of moles of hydro? Well, we're given the concentration of 0.50 molar from HCL and we're taking some volume, let's call it V one. Whenever we want to identify moles, we multiply concentration by volume as seen in the formula, right. So this first part gives us the number of moles of hydro num coming from HCL. And now we are adding the number of moles of hydro coming from HNO three. So we have to take the concentration of 0.30 molar and multiply by the same volume V 11 mixing equal volumes. And on the bottom, the total volume is just V one plus V one because the volumes are additive. And now if we factor V one out, we get V one divided by two V two. And now in pres we just have 0.50 molar plus 0.30 molar when we simplify the equation, we can see that we can cancel out V one. And it essentially tells us that it doesn't matter what the volume is. We don't really worry about the value of V one. It's essentially 0.80 divided by two, the total concentration of 0.40 molar of hydro. And therefore, we know that ph is equal to negative log of the concentration of hydro, which is 0.40 molar. If we evaluate the result, we get 0.40. Therefore, the correct answer to this problem is 0.40. That's the Ph of the resultant solution. Thank you for watching.