What is the pH of a buffer made with 0.273 M nitrous acid...

What is the pH of a buffer made with 0.273 M nitrous acid (Ka = 6.91 x 10^-4) and 0.288 M sodium nitrite? 10.9 3.13 3.19 0.524 13.4 Question 2 Which of the following constitutes a pH buffer system? 0.1 M HNO3 and 0.1 M NaNO3 0.1 M CH3COOH and 0.1 M NaOH 0.1 M NH3 and 0.1 M CaCl2 0.1 M NaClO and 0.1 M HClO 0.1 M HBr and 0.1 M KOH Question 3 What is the pH of 100.0 mL of a buffer made with 0.140 M each of chloroacetic acid (CH2ClCOOH, pKa = 2.85) and sodium chloroacetate (NaCH2ClCOO) after the addition of 15.1 mL of 0.449 M HCl? 2.56 2.85 3.14 2.36 1.54

Transcript

00:01 Hi, so over here we want to answer the following questions regarding ph of weak acids and weak bases and buffer systems.

00:07 So over here, first of all, what is going to be the ph of a buffer with the following concentrations of nitrous acid? this is going to be our weak acid and sodium nitrite, which is going to be our conjugate base.

00:19 So in this case the ph is going to be given by the henderson -hasselbalch equation which tells us that ph is equal to pka plus the logarithm of the concentration of the conjugate base divided by the concentration of the weak asset.

00:33 The pka for this asset is equal to minus the logarithm of ka, giving us a pka of 3 .16.

00:42 So replacing the values we're going to get 3 .16 and let me put it over here 3 .16 plus logarithm of 0 .288 over 0 .273 in this case we're going to have more of the conjugate base than of the acid so we're going to expect the ph to be higher than the pka so the result over here is going to being equal to 3 .18 as a ph well 3 .1834 so this would be our final answer sorry about it in this case since you changed the values probably these sensors make no sense so it would be 3 .183 as the final ph now which of the following constitutes a ph buffer system for a buffer system we're going going to have a weak acid with its conjugate base.

01:58 So the systems that we're going to have are going to be nitric acid with nitrate.

02:04 This is not going to be a buffer system since this is going to be a strong acid.

02:09 We're going to have acetate with acetic acid with selenium hydroxide, also not a buffer system.

02:16 Nh3 with calcium chloride, not a buffer system.

02:20 Hbr is a strong acid with koh also not a buffer and the only one that is going to be a buffer is the nacllohclo since these are going to be a weak acid with its conjugate base present in solution and finally what is going to be the ph of 100 milliliters of a buffer made with 0 .140 molar each of chloroacetic acid and and sodium chloroacetate, which the pka is going to be provided.

02:52 After the addition of 15 .1 milliliters of 0 .449 molar hcl.

02:59 So the amount of mole of h3o plus that we are adding to the system is going to be equal to the volume of the acid in liters times the concentration in molarity of the acid.

03:10 The volume added in this case would be 15 .1 times 10 to the minus 3 liters.

03:19 Times 0 .449 mol per liter.

03:24 This amount of mol is equal to 6 .7799 times 10 to the minus 3 mol...

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