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: equivalence point.

Accepted Answer

Hi everyone here we have a question that says at the equivalence point, what is the ph of the thai tradition of 20 mL of 0.178 molar acetic acid with 0.150 molar sodium hydroxide. So at the equivalence point the molds of acid is going to equal the molds of the base and all of the bases used up and the acid is converted to its consequent base. So our step one is to determine the volume of base needed to reach the equivalence point. So we have 20 ml times one liter, Divided by 1000 ml times zero . Mnolds over one liter. And our middle leaders are canceling out and our leaders are canceling out and that gives us 3. Times 10 to the negative 3rd moles of acetic acid. And now we have our 3. Times 10 to the -3 moles of acetic acid times one more of sodium hydroxide over one mole of acetic acid times one leader Divided by 0.150 moles of sodium hydroxide. And our mole's of acetic acid are canceling out and our moles of sodium hydroxide are canceling out, leaving us with 0. 73 L. So now we need to set up an ice chart. So we have acetic acid plus hydroxide goes back and forth between its conjugate base, which we're going to call a C. O plus water. And our water isn't going to matter because it is a liquid, It's not acquis and we're going to have our initial our change and our end. So our initial is going to be 3. times 10 to the -3 moles, 3.56 times 10 to the negative three and zero. And our change is going to be negative 3.56 times 10 to the negative three And negative 3.56 Times 10 to the -3 And plus 3.56 Times 10 to the -3. So our end is going to be 00 and 3.56 times 10 to the -3. Our total volume of the solution at the equivalence point equals 20 mill leaders Plus 23.73 ml Divided by 1000 Which equals 0. 373 L. Our concentration of our conjugate base Equals 3.56 Times 10 to the -3 moles Divided by 0. 4373 L, which equals 0. 8141. Now we need to do an ice chart for the conjugate base of acetic acid. So we have our conjugate base plus water goes back and forth from acetic acid plus hydroxide. And we're going to have our initial or change and our end Our initial is going to be 0.08141. Our water doesn't matter. Our initial is going to be zero and zero. Our change is going to be minus X plus X. And plus X. Our end is going to be 0.0814, one minus x. X. And X. So now our kB equals kw divided by K. A. Equals one times 10 To the -14, Divided by 6.4 Times 10 to the -5 equals 1. times 10 to the negative 10. R K B equals X times X, Divided by 0. minus X Equals 1.5625 times 10 to the -10. So our concentration of H. A. Divided by K. A. Equals zero point 08141 Divided by 1.56, times 10 To the negative 10 And that is greater than 500. So X. can be neglected. So X squared equals the square root of one .5625 Times to the negative times 0. equals 3. times 10 to the sixth. So our concentration of hydroxide Equals 3. 55 times 10 to the -6. And lastly our P. O. H. Equals our negative log of. Our concentration of O. H equals the negative log of three . times 10 to the - equals five .44775. R P h equals 14 minus R P. o h. So that equals 8.55. And that is our final answer. Thank you for watching bye.

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: equivalence point.

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