Important equations pH = pKa + log[salt]/[acid] acidic pKa...

Important equations pH = pKa + log[salt]/[acid] acidic pKa = - log Ka pOH = pKb + log[salt]/[base] basic pKb = - log Kb Acid Ka Sodium salt Acetic acid CH3COOH 1.8 x 10-5 NaCH3COO Benzoic acid C6H5COOH 6.3 x 10-5 NaC6H5COO Hydrocyanic acid HCN 6.2 x 10-10 NaCN Formic acid HCOOH 1.8 x 10-4 NaHCOO Hypobromous acid HOBr 2.3 x 10-9 NaOBr Hypochlorous acid HOCl 2.9 x 10-8 NaOCl Hypoiodous acid HOI 2.3 x 10-11 NaOI Nitrous acid HNO2 7.1 x 10-4 NaNO2 Chlorous acid HClO2 1.1 x 10-2 NaClO2 Solve the following problems. 1. Find the pH of a buffer in which [HCOOH] = 1.24 M and [HCOONa] = 0.82 M 2. Given that Buffer A contains 200.0 mL 0.050 M HOCl and 400.0 mL 0.030 M NaOCl, calculate a. The pH of the buffer solution. b. The pH of the solution after adding 10.0 mL 0.50 M HCl. c. The pH of the solution after adding 20.0 mL 0.40 M NaOH

Transcript

00:01 For this problem we have a buffer that is made up of hcooh, which is an acid, and hcoona, which is the conjugate base of that, and we're asked to find the ph.

00:15 So our equation for a buffer is h plus concentration equals ka times our acid concentration over the conjugate, like weak acid over its conjugate base or conjugate acid over the weak base.

00:30 So our ka here is 1 .8 times 10 to the negative fifth, excuse me that's to the negative fourth, so 1 .8 times 10 to the negative fourth.

00:44 Our acid concentration, that's this right here, is 1 .24 molar and our base concentration is 0 .82.

00:52 So we simply plug this in, solve for our h plus concentration, so 1 .8 times 10 to the negative fourth times 1 .24 divided by 0 .82 equals 2 .72 times 10 to the negative fourth molar, and then we simply take the negative log of that to find our ph, and so here our ph would equal 3 .57.

01:22 Okay, the next thing we have is a combination of different things going on here.

01:27 We have first wanting to know the ph of a buffer made of these two components, and so we need to find our concentration of this.

01:42 Our equation is going to be the same, so h plus equals, so we're going to take our 200 times 0 .05, 200 times 0 .05 divided by the total volume, which is 600 ml, which is going to give you point, oops i need my ka, so my ka is 2 .9 times 10 to the negative 8.

02:08 Sorry, i got ahead of myself there.

02:10 Okay, so i took the 200 times 0 .05 divided by the total volume of 600, and that gave me a concentration 0 .0167 molar, and then for the bottom part it is the 400 times 0 .03 divided by the 600 ml total, which is 0 .02.

02:35 So now 2 .9 times 10 to the negative 8 times 0 .0167 divided by 0 .02, and that will give you 2 .42 times 10 to the negative 8, and then you will take the negative log of that to find your ph, which is 7 .ph equals 7 .62.

03:10 We'll now continue to look at this when we add hydrochloric acid and sodium hydroxide in two separate pieces here.

03:21 So our h plus concentration equals 2 .9 times 10 to the negative 8...

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