Post Lab Problem 1 (5 points). A buffer is made by mixing 10...

Post Lab Problem 1 (5 points). A buffer is made by mixing 10 mL of 3.0 M HCOOH (aq) with 5.0 mL of 1.44 M KHCOO (aq). What is the pH of the buffer at 25°C, if the acid ionization constant for formic acid is $1.7 \times 10^{-4}$? Round off your answer to 3 sig. figs. and enter it in the box provided with correct units and sig. figs.: Comment: Henderson-Hasselbalch equations for acidic and basic buffers are listed on the previous page. Answer: $pH_{buffer}$ =

Transcript

00:01 And this question 3 difference it has been given and we are provided with the following value that is 0 .25 liters of buffer solution the molar concentration of hf and nf has been given.

00:10 So, let's proceed with the first one we need to calculate the ph of the buffer solution.

00:15 So, for calculating the ph of the buffer solution we will use the henderson -helschbach equation which is equal to ph is equal to pka plus log of a minus divided by the ha.

00:47 So, now what is pka? we know that pka is the acid dissociation constant, a is the concentration of the conjugate base in this case we can say it is for nf and ha is the concentration of the acid in this case it is for hf.

01:08 So, first we need to find it out the concentration for naf and hf.

01:12 So, the concentration for naf would be equal to what that is the molarity into concentration that is the molarity that is equal to concentration into the volume.

01:25 So, the concentration has the molarity for nf is 0 .35 molar concentration into the volume that is 2 .50 liter.

01:35 So, the concentration of nf is come out to be 0 .0875 mole.

01:43 Next let's find out the concentration for hf the same would be there because the molarity or the concentration is same for both.

01:52 So, it would be equal to molarity into volume again that is 0 .35 into volumes are 2 .50 liter.

02:00 So, i am calculating it comes out to be 0 .8075 mole.

02:06 Now, we need to calculate the concentration of the a by ha.

02:10 So, we can put the values into it or we can directly put the formula.

02:15 So, our ph that would be equal to ph would be equal to the pka that was equal to pka plus log of a minus by ha.

02:33 So, the value of pka has been provided to us that is equal to 3 .77 plus log of 0 .0875 that is the concentration of a divided by hf that is equal to 0 .0875.

02:58 So, i am calculating this the ph for the first part it comes out to be 3 .77.

03:06 Let's move to the second question or the second part.

03:11 In this question we have to calculate the ph of the buffer solution, ph of the buffer solution after addition of 0 .0025 mole of hno3.

03:33 So, we need to again use the henderson -hasselbalch equation which is equal to ph equals to pka plus log of a minus divided by ha.

03:48 So, again we have to add the concentration of ha, h plus hay because this much amount of acid has been added.

03:57 So, addition of h plus divided by ha...

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