A buffer solution is prepared by dissolving 0.634 moles of...

A buffer solution is prepared by dissolving 0.634 moles of nitrous acid (HNO2) and 0.191 moles of the salt sodium nitrite (NaNO2) in 1 L of solution. a. Use the Henderson-Hasselbach equation to compute the pH of this buffer solution. b. 0.055 moles of the strong base NaOH is added to the buffer solution described above. Assume that the addition of this strong acid does not change the volume of the solution. Use ICF and ICE tables to compute the concentration of H+ and pH after the addition of the base to the buffer solution. Be sure to show your work. c. 0.037 moles of the strong acid HI is added to the buffer solution described above. Assume that the addition of this strong acid does not change the volume of the solution. Use ICF and ICE tables to compute the concentration of H+ and pH after the addition of the base to the buffer solution. Be sure to show your work.

Transcript

00:01 All right, we're told that a buffer solution is prepared by dissolving 0 .63 moles of nitrous acid and 0 .19 moles of sodium nitrite, the salt, which in a one liter solution.

00:14 So basically, the concentration of the nitrous acid will be the number of moles divided by the volume of the solution, which is one, so that's 0 .634 molar.

00:26 On the other hand, the concentration of the conjugate base of the acid, that's the nitrous nitrite ion, would also be the number of moles divided by the volume, so that's 0 .191 molar.

00:41 Now the ph of the solution using henderson -hasselbalch is pka of the nitrous acid plus the log of concentration of the base divided by the concentration of the acid.

00:57 Now the ka of nitrous acid is 4 .5 times 10 to the power of minus 4, and so the pka is basically negative log of that, of the ka, and that should give you 3 .76, no, 3 .35 actually, 3 .35 as a pka.

01:22 So we're going to put that in, so the ph would be 3 .35 plus the log of 0 .191 divided by 0 .634, and when you punch your calculators, this would give you 2 .83, right, that is the ph of the buffer.

01:48 Now let's look at the next what happens, right, the b part says what happens when you add a certain amount of moles of a strong base to the buffer, what happens? now we're going to start by using the icf tip, which is just initial constant change in and then what is formed.

02:13 So here's it, when you have a base, sodium hydroxide, a strong base, a sodium hydroxide react, this is what happens, the reaction goes, you have the weak acid would react with this, you would have the conjugate base and then water, right.

02:34 So initially, initially we had 0 .634 molar of this, now since the volume of the buffer doesn't change, we can say that the concentration of the hydroxide ion is also the same as the number of moles of hydroxide ion added, since we had just one liter, so 0 .055 molar and this was 0 .191, now this doesn't count.

03:02 Now the change is, so all of the base would react, right, all of the base would react and of course this will be, this is added, so this is minus, this is minus, this is going to be plus.

03:23 So what is formed and left behind at the end of the day, this the acid is still in excess by 0 .0579 and this is 0 .691.

03:41 This would be the new concentrations of the weak acid and its conjugate base in the buffer.

03:49 Now the ice, but then we want to know the concentration of the hydrogen ions or the hydronium ion and so we're still going to need to set up, right, so if we had this, water ionizes and you have this and hydronium ion or hydrogen ions, so now the initial, this is 0 .059, 579 and this was 0 .691.

04:20 Change some of these to the lost, that would be gained here and that would be gained here.

04:25 So we want the equilibrium, at equilibrium 0 .0579 minus x, this is x and that is, oh sorry, this would be 0 .691 plus x.

04:39 The ka value is what we're going to be using now, the ka of the nitrous acid and the ka value like i said earlier is 4 .5 exponential minus 4, it's quite small.

04:53 So this approximately is actually 0 .0579 and this is also approximately 0 .691.

05:01 So the ka, the ka is the concentration of the nitrite ion times that of the hydrogen ion or the hydronium ion divided by the concentration of the base, of the acid, sorry, not the base.

05:17 So basically we want the concentration of the hydronium ion which would be the ka, 4 .5 times 10 to the power minus multiplied by 0 .579 divided by 0 .691 and this would give you 3 .77 times 10 to the power minus 4.

05:38 That is the concentration of the hydronium ion and the ph will be what? negative log of that and that would give you definitely 3 .4, 3 .42.

06:00 Now you can see that the ph has increased, right? the ph has increased from this.

06:05 That's the effect of adding a base, right? you increase the ph because you're decreasing the concentration of the weak acid and increasing the concentration of the conjugate base of the acid.

06:15 Okay so the second, the other question would be the same, right? the last one.

06:19 When you add an acid it would also follow a similar path.

06:23 This time around when you add an acid you are going to affect the conjugate base.

06:29 So you would have the no2 would react with the acid and you would form the conjugate base or the acid, the conjugate acid and water...

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