A 30.0-mL sample of 0.165 M propanoic acid is titrated with
0.300...

Question

A 30.0-mL sample of 0.165 M propanoic acid is titrated with
0.300 M KOH. Calculate the pH at each volume of added base:
20 mL.

Accepted Answer

Hi everyone here we have a question asking what is the ph of the solution of mL of 0.158 moller nitrous acids high traded with 30 mL of 0.145 molar sodium hydroxide. So first we're gonna convert this to moles. So we have 25 mill leaders and we need to change that to leaders. So we're gonna multiply by one leader Over 1000 ml and then we're going to multiply that By 0.158 moles divided by one leader. So our milliliters are going to cancel out and our leaders are going to cancel out and that's going to give us 3. Times 10 to the -3 moles of nitrous acid. And then we're gonna do the same thing with our 30 ml of sodium hydroxide. So we have 30 ml. We need to change that to a leader. So we're gonna multiply by one leader Over ml and we're going to multiply that by 0.145 moles over one liter. And our mill leaders are going to cancel out and our leaders are going to cancel out, giving us 4.35 moles, 4.35 times 10 to the - moles of sodium hydroxide. So we're going to let H a stand for our nitrous acid and we're gonna let a minus stand for its conjugate base. So we have h A plus O H minus goes back and forth to a minus plus water. And we're going to do our initial a change and our end. So initially we're starting off with 3. Times 10 to the negative 3rd moles of nitrous acid And 4. Times 10 to the -3 moles of hydroxide. And zero of our conjugate base. And our water isn't going to matter because it is a liquid, it's not acquis. So our nitrous acid, we're going to have negative 3. Times 10 to the -3 and same for our hydroxide. And for our conjugate base we're going to have plus 3.95 Times 10 to the -3. So for our ending we have zero four times 10 to the -4 And plus 3.95 Times 10 to the -3. Because sodium hydroxide is a strong base, potassium hydroxide is an excess which gives us hydroxide. The strong base overwhelms the weak base which is significant in the calculation. So our total volume is 25 plus 30. She equals 55 ml. And we're going to convert that to leaders. So Times one leader Over ml Our male leaders cancel out And that gives us 0. L. Our concentration of hydroxide Equals four times To the -4 Divided by 0.055 l Equals 7.2727 times 10 to the -3. Our P. O. H equals the negative log of 7. times 10 to the - Equals 2.1383. RPH equals minus r p o h equals -2. Equals 11.86. And that is our final answer. Thank you for watching. Bye.

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