A 35.0-mL sample of 0.150 M acetic acid 1CH3COOH2 is titrated
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Question

A 35.0-mL sample of 0.150 M acetic acid 1CH3COOH2 is titrated 
with 0.150 M NaOH solution. Calculate the pH after 
the following volumes of base have been added: (b) 17.5 mL.

Accepted Answer

Hello everyone today. We have the following problem a 0.250 moller potassium hydroxide solution was used to titrate a 50 million solution of 500.155 bowler of hipaa Cloris acid. And it has the following K. A. Value there determine the resulting ph of the solution. After 14.5 mL of potassium hydroxide has been added. So the very first thing that we want to do is we want to calculate the initial number of moles. You want to calculate the initial moles for both our hipaa Cloris acid and are potassium hydroxide. So first we will work with our hipaa Cloris acid. We will start with what we're given which is our 50 ml. And of course we have to convert this to leaders. So we use the conversion factor that one millimeter is equal to 10 to the negative third leaders. And then we multiply by the polarity similarities and units of moles per liter. So this is simply .155 moles over one liter units of liters and milliliters canceled out. And we are left with 7.75 times 10 to the negative third moles. And we'll hang on to that number for later we then need to find our number of moles for potassium hydroxide. Once again we use the same method which is our volume not 14. ml. We have to convert these two leaders. So we use the conversion factor that one millimeter is equal to 10 to the negative third leaders. And then we multiply by the malaria T, which in his case is going to be .250 moles per one leader. And the resulting answer is 3.625 times 10 to the -3 moles. Well we use that number for later. The second thing you wanna do is you want to write the dissociation for potassium hydroxide. So that essentially means that when we have potassium hydroxide it breaks into potassium ions as well as hydroxide, Aquarius ions. And so hydroxide and this situation is going to be our base and this is just gonna be a neutral ion. Our potassium will be a neutral ion. So the third thing we wanna do is you wanna draw I cf table our I. C. F. Table which is essentially what occurs when we have our hipaa Cloris acid and we titrate it with our potassium hydroxide. We're going to form Our Quarry eight ion as well as water. And so by I. C. F. We have our initial concentration the change and the final. And so we calculated the initial concentration of a hydrochloric acid which is a 7.75 times 10 to the negative third. And for potassium hydroxide that is 3.625 times 10 to the negative third. Of course we don't have one for the chloride ion and water will not have any play or any role in this situation. And so since we are using potassium hydroxide to titrate it. We're simply going to say that we have a change of minus 3.625 times 10 to the negative three for both of our reactant. And so When we calculate the change or the final, which is the initial minus the change. That's going to be 4.125 times 10 to the negative third moles. And if the potassium hydroxide of course that will be zero. And then our chloride ion will be adding that 3.625 times 10 to the negative third. So of course the final is going to be that zero plus that 3.625 times 10 to the negative third. Giving us this answer and so forth. Now that we have our values, we want to make note of acidity and basis city. So we have our hipaa Cloris acid here and this is going to be a weak acid. So it's not going to completely dissociate just partially. And then we have our chlorate ion here which is going to be a weak base because it is coming from a weak acid. So a strong base, a strong conjugate base would come from a strong acid and a weak conjugate base would come from a weak acid. And so now that we've solved that we need to find our total volume, which would be known denoted by V sub T. That is going to be using our 50 middle leaders of course, changing this to Leaders by using the conversion factor from earlier And they were also going to add that to our 14.5 ml which will also be converted Into leaders. And when we do that we get a final volume of . 45 L. And we're gonna hang on to that number for later. Our 6th step is going to be defined our concentration of our ions first. We're gonna use our hyper Cloris acid hcl oh and that's going to be found by taking our moles that we sold for which is our 4.125. Our final concentration of moles times 10 to the negative third over Our volume that we just sold for our .0645 l To give us 0. moller. And then for chlorate we are simply going to do the same procedure. This is going to be 3.625 times 10 to the negative third moles over the volume we sold for which is 100.645 liters giving us 0.56 to zero molar. So we're going to hang onto these two numbers as well. Our seventh step there's 10 in total is to write out our K. A expression. R. K expression is simply going to be the concentration of our products. Overreacting so and this is going to be for our initial reaction that we had which is this one that we did our I. C. F. Table for. So for products we only include acquis solutions, we do not include liquids. So that's just going to be our cielo minus. And instead of water we're going to include hydroxide for our reactant, we're simply just gonna have the hipaa Cloris acid. We will not include our potassium hydroxide and so we simply essentially just have to solve for H 30. Plus or hydro um ion because once we plug in our numbers here for example, RK according to the question still was three times 10 to the negative eighth equals the concentration from arklow rate which is 0.56 to zero molar. We don't have the concentration for our hydro ni. Um that's what we need to find over our concentration for hyper Cloris, for our Hippo Florist asset which is zero 06395 Moller. And when we solve for that we get a hydro me um concentration of 3.415 times 10 to the -8 moller. And so our last step I convinced a couple of steps, our last step will be step eight. We're going to solve for P. H. Which the formula for that is a negative log times our concentration of our high germanium ions. And so that's simply negative log of what we just found. Our 3.415 times 10 to the negative eight molar. And when we sell for that in our calculator we get a ph of 7.467 as our final answer. Overall, I hope that this helped until next time.

A 35.0-mL sample of 0.150 M acetic acid 1CH3COOH2 is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added: (b) 17.5 mL.

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