Calculate the pH of 0.375 L of a 0.18 M acetic acid–0.29 M sodium...

Question

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 (b) 0.0060 mol of HBr. Assume that the volume remains constant.

Accepted Answer

Welcome back everyone. We need to determine the ph of a 0.500 liter solution and a 0.30 moller lactic acid 0.25 molar sodium lactate buffer before and after 0.90 moles of hydrochloric acid is added. Assume the volume of sodium hydroxide is negligible and the PK of lactic acid is 3.86. So we're given that we have 0.30 moller of lactic acid. Recall that the formula for lactic acid is going to be a church C. H. Three H 503. And then we have a point Molar sodium Lactate. So recall that our sodium lactate being a salt is going to disassociate from the falling formula. We have N. A. C. H. Three H 5031 minus which dissociates fully when it dissolves in water to a sodium catalon plus C. H. Three H +5031 minus being our lactate an ion. And so our .25 Molar sodium lactate also equals our concentration of our lactate an ion and we should fill in our labels since these are all Aquarius. So we need to calculate first our ph before our hydrochloric acid is added for the solution, recall the formula where our ph is equal to our P. K. A. Of our weak acid lactic acid added to the log of the concentration between our conjugate base a minus divided by our concentration of our conjugate acid H. A. So plugging in our knowns, we would find that our ph before hydrochloric acid is added Ph is equal to R. P. K. of lactic acid given in the prompt as 3.86. This is added to our log and I'll make some room here. Our log between our concentration of our conjugate base being our lactate an ion because it's missing that proton in the formula. So we have the concentration of C H three H minus Divided by our concentration of our conjugate acid being our lactic acid, HCH three H 503. And so plugging in our concentrations we would have ph equal to 3.86 plus the log of 0.25 moller divided by our denominator. Our concentration of lactic acid 0.30 moller for our conjugate acid. This will simplify to a ph before hydrochloric acid is added. For the solution equal to about 3.78. So now for our next step which is to find our ph after hydrochloric acid is added. We're going to have to take a few more steps where we're gonna first need our moles of our conjugate base and acid and conjugate acid. Sorry, because our hydrochloric acid is going to react with our conjugate base or lactate an ion. So recognize that hcl is a strong acid and so it's going to fully dissociate into our chloride one minus an ion and an H plus proton. And so that means that we would have hpe lus reacting with our lactate an ion C H three H +5031 minus. And this is how our lactic acid is formed. H c H three H 503 as a product. So we're going to move this equation down so that we can first find our moles of our conjugate acid and base. So now we have space. We want to find most of our lactate an ion CH three H 503 minus. We're going to begin with it's Mueller concentration given in the prompt as 5030.25 Mueller for sodium lactate, which it associates into our lactate an ion. So that concentration is equivalent. And instead of our polarity unit we're going to recall that polarity is equivalent to moles per liter. So we will interpret this concentration in terms of moles per liter. So we want molds to be our final unit. And we're going to multiply by the volume of our solution given in the prompt as 0.500 leaders. Which will allow us to cancel out leaders leaving us with moles which is what we want. And in our calculators will find that we have 0. 125 moles of our lactate an ion. And now for our moles of our lactic acid H c H three H 503 we have from the prompt. Its molar concentration given as 0.30 moles or Mueller which we are interpreting as moles per liter. And again multiplying by the volume of our solution which is still 0.500 liters allowing us to cancel out leaders leaving us with moles in our numerator as our final unit will find in our calculators. 0.15 moles of our lactic acid. So these are going to be our initial values for our ice chart, which is what we are moving to next in our next step. So plugging in the initial concentration of our hydrochloric acid given in the prompt as 0.90, which will also be the initial concentration of our proton H Plus. Since hydrochloric acid will dissociate into H plus We then want to plug in our calculated initial concentration for our lactate an ion which we determined to be .125 moles. And just for room sake we're not going to write animals. And then for our lactic acid we calculated an initial concentration of 0.15 moles. So now we want to recall that from the change portion of our ice table. We would use the smallest value and we can see our smallest value is our initial concentration of H plus from hydrochloric acid. So we would subtract this amount from our reactant minus 0.90 since H plus is limiting. And then we would add this amount plus 0.90 on the product side. So for our equilibrium we're just gonna carry the math down here. So we would have zero left of hcl or H. Plus. Rather we would have for our lack date and ion a difference of 0. moles. And then for our lactic acid we would have a sum of 0.159 moles. So now we just need to interpret these equilibrium concentrations in terms of polarity again. So now for our concentration of our lactic acid, H ch three H We would take the moles that we just found .159 moles and recall that we divide this by our volume of our solution given in the prompt as 0.500 L. So we have moles per liter which is what we want for units of polarity. And in our calculators this will equal a concentration of 0.318 molar for our lactic acid. And next we want our concentration of our Conjugate base lactate. An ion where we have CH three H 5031 -. We determined its equilibrium concentration to be 0. moles divided by the volume of our solution given in the prompt as 0. liters. We will have a result of this quotient in our calculators as 0. 32 moller. And so now going back to our ph formula, we have the ph equal to R. P. K. A. Of lactic acid plus our log of our concentration between our conjugate base and our conjugate acid, H. A. Plugging in our values. We have again our P. K. A. Of lactic acid given us 3.86 plus our log of our concentration of our conjugate base which we determined based on our calculation above To be .232 molar. This is divided by our concentration of conjugate acid being lactic acid, which we determined to be .318 moller Our units of polarity cancel out and what we would have is a ph equal to about 3.72. And this would be the end of our example. So we have our ph before hydrochloric acid is added to our solution being 3.78. And then we have our ph after hydrochloric acid is added, being a value of 3.72. And so our two answers highlighted in yellow will correspond to choice D in the multiple choice as the correct answer. So I hope that this made sense. And let us know if you have any questions

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 (b) 0.0060 mol of HBr. Assume that the volume remains constant.

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