Consider the titration of 40.0 mL of 0.250 M HF with 0.200 M NaOH...

Question

Consider the titration of 40.0 mL of 0.250 M HF with 0.200 M NaOH. How many milliliters of base are required to reach the equivalence point? Calculate the pH at each of the following points. 
(d) After the addition of 80.0 mL of base

Accepted Answer

Hello in this problem we are told 35 mL of a 350.225 Mueller satanic acid was tight rated with 0.215 miller potassium hydroxide were asked to determine the volume of testing hydroxide necessary to reach the equivalent point and to calculate the ph after the addition of 85 mL of base for given the PK for scion IQ acid as 2.46 km. Let's begin by writing the chemical reaction. We have sonic acid, then we'll react to the potassium hydroxide, we will form water and our assault potassium sign eight. Next step is to calculate the volume to the equivalence point. So we have 35 ml. It's ionic acid convert this leaders And later as you go to the 1000 ml concentration. Then used to pervert for volume to moles Those . moles. Sonic acid for every leader. And based on the psyche geometry from a balanced reaction equation will model of potassium hydroxide reacts with one mole of sonic acid. Can you Some concentration of potassium hydroxide. There are .25215 moles of potassium hydroxide. Every leader that solution And we'll convert two ml One later is equal to L. And so our units, the volume of sonic acid cancels. Leaders, sanick acid cancels moles sonic acid cancels moles of potassium hydroxide cancels leaders of potassium hydroxide cancels and we're left with the leaders of potassium hydroxide which works out to 36. male leaders of potassium hydroxide to reach the equivalence point. And then our next step, let's now calculate the moles of tannic acid and molds of calcium hydroxide. So we have and 35 mm of ultrasonic acid. Following the steps that we did above convert two mL milliliters to liters and then we'll make use of similarity to find moles. So our unison, the leaders of cynic acid cancels. Leaders of cynic acid cancels. And this works out to 0.00 walls of our acid. Now we'll calculate the molds of our base. So we Added 85 ml of gas and hydroxide convert our volume four mL to liters. Make use of its concentration Which is 0. per liter. The leaders of potassium drugs, I cancel leaders of potassium hydroxide cancel. And this works out to zero point 018 275. Those of our face on our next step. Then we're going to find moles of our acid, most of our contract base and those of hydroxide. After the addition of testing hydroxide. So we rewrite our reaction. So we have initial change and then final. So initially have 0.007875 moles of our acid, 0.018275 Roles of our base. And initially the water or salt. So we're going to assume then that all of our acid reacts -0.007875. And so we have no acid left. Yeah 0.0104 modes of our strong base left over. Which means that we've passed the equivalence point. And that means then the ph after the addition of 85 ml of potassium hydroxide will depend on the not a strong base. And then our next step will calculate the th so our P. O. H. Then will be the negative log of our hydroxide ion concentration. And so we have 0.0104 moles our base the total volume. Then so we added milliliters of acid and 85 ml of base. So that's our combined volume, convert our volume. Then from male Leaders, Two leaders simulators cancels and we're left with a unit of concentration. So our P. O. H works out to 1.62. And our ph sen we'll take 14 minus the p O h 14 -1.062. And this works out to then 12.938. So we are finding the ph we have just strong base left. So this ph value makes sense. And we were also again, In addition to finding the ph after edition of 85 mm, we're supposed to find the volume at the equivalence point. Again, it took us 36.6 mL of potassium hydroxide to reach the equivalence point. So with the addition of 85 mL. We've passed the equivalence point. And we'll have excess strong base left over. Which then accounts for our high ph value. And this then corresponds to answer D. Thanks for watching. Hope. This helps.

Consider the titration of 40.0 mL of 0.250 M HF with 0.200 M NaOH. How many milliliters of base are required to reach the equivalence point? Calculate the pH at each of the following points. (d) After the addition of 80.0 mL of base

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