Calculate the pH of 0.375 L of a 0.18 M acetic acid–0.29 M sodium acetate buffer before and after the addition of (a) 0.0060 mol of KOH. Assume that the volume remains constant.

Video transcript

Hello everyone today we are being given the following problem determine the ph of a 0. liter solution of or that is composed of 0.15 miller lactic acid and 0.25 molar sodium lactate buffer. Before and after 0.0 50 moles of sodium hydroxide is added, assume the volume of acts is negligible. The PK of lactic acid is 3.86. So first you want to recall our Henderson household back equation so that we can determine the ph before edition. So the Henderson Hasselbach equation states at the ph is equal to the P. K. Plus the log of our conjugate base divided by our concentration of our asset. And so the P. K. We were given was 3.86 for our lactic acid. And then we have the log of our conjugate base which was 0.25 or the lactate. And then our asset of course is .15. So this gives us a ph of 4.06. So the ph before edition Is 4.08. So now we need to find a ph after the addition. And so the sodium hydroxide is going to react with that acid. And we have to determine the moles of each first Before we get to the ph so we're gonna find the moles initially of our lactic acid Or C3 H 603. And so to do that we're gonna take our polarity and multiplied by the volume similarity is in terms of moles per liter. So we're gonna have 0.15 moles of lactic acid divided by one liter, multiplied by the solution which was 10.500 liters. And when our units cancel we get 0.75 moles. And then for our molds of our like Tate which is going to be C three H minus. We're gonna use the same Process. We're gonna take the allergy .25 moles and divided by one leader and then multiplied by the .500 L to give us .125 moles. So we have the moles of our acid and our conjugate base. So now we need to find or solve an ice table. And so that's going to be our lactic acid. We're just gonna have a formula of C3 H603. And of course we're going to react it with our hydroxide. This is going to form our lactate. They're gonna have our initial concentration are change and our equilibrium or the final. So we started out with 0.075 moles of our lactic acid and 0.050 moles of our hydroxide that we used. And this ultimately gave us 0. moles of our conjugate base. So since we're tight trading with our bases are hydroxide, we're gonna subtract by that number. So we're gonna subtract everything on the reactive side, Bit .05. And we're gonna add it to the products. E. Is just simply the I. And the sea rose added up. It's gonna give us .07 moles. We're gonna zero for hydroxide of course. And then we're gonna have .13 moles for our conjugate base. And so with this we're going to find our concentrations of both. So first with our concentrations of our lactic acid, it's going to be our 0.07 moles divided by our volume 0.5 liters to give us 0.14 moller. And our concentration of our lactate Is going to be our .13 moles divided by 0.5 L. And that's gonna give us 0.26 molar. So now we have our concentrations of our acid and conjugate base. We can plug this into our secondary Henderson household buck equation here and that's gonna be the ph once again, as they go to the PK plus our log of our concentration of our conjugate base divided by the concentration of our acid. And so we have our PK once again missed 3.86 Plus our log and our concentrations are going to be .26 for our conjugate base and 0.14 four hour acid playing this in the calculator, we will get 4.13. So our ph after Adding that hydroxide, it's going to be 4.13. And so with that we have our final answers overall, I hope this helped. And until next time