Use information from Appendix D to calculate the pH of (a) a...

Use information from Appendix D to calculate the pH of (a) a solution that is $0.250 M$ in sodium formate (HCOONa) and $0.150 \mathrm{M}$ in formic acid $(\mathrm{HCOOH}),(\mathbf{b})$ a solution that is $0.490 \mathrm{M}$ in pyridine $\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)$ and $0.450 \mathrm{M}$ in pyridinium chloride $\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHCl}\right)$, (c) a solution that is made by combining $80 \mathrm{~mL}$ of $0.050 \mathrm{M}$ hydrofluoric acid with $100 \mathrm{~mL}$ of $0.10 M$ sodium fluoride.

Key Concepts

Buffer Solution

A buffer solution is a mixture that resists changes in pH when small amounts of acid or base are added. Typically, it consists of a weak acid and its conjugate base or a weak base and its conjugate acid, which work together to neutralize added H+ or OH- ions, maintaining a relatively stable pH.

Conjugate Acid-Base Pair

A conjugate acid-base pair is formed when a weak acid donates a proton to become its conjugate base, or when a weak base accepts a proton to form its conjugate acid. These pairs are central to buffer behavior, as the relative concentrations of the acid and base determine the pH of the solution.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a formula that relates the pH of a buffer solution to the pKa of the weak acid and the ratio of the concentrations of its conjugate base to the acid. It provides a straightforward method to calculate the pH of buffer solutions by using the equation pH = pKa + log([Base]/[Acid]).

Acid Dissociation Constant (Ka) and pKa

The acid dissociation constant (Ka) quantifies the strength of a weak acid by measuring its degree of ionization in water. The pKa, which is the negative logarithm of Ka, provides a more convenient numerical value for comparing acid strengths and is an essential parameter in buffer calculations.

Molarity and Dilution

Molarity is a measure of solution concentration defined as moles of solute per liter of solution. When solutions are mixed, especially in the formation of buffers, the molarity of the components may change due to the combined volumes. Accurately calculating these new concentrations is crucial for correctly applying the Henderson-Hasselbalch equation in pH determinations.

Transcript

00:01 This is a long problem, but not very hard.

00:04 We have three problems to solve.

00:05 I'll do them each separately.

00:08 For our first problem, we're given the following information.

00:11 We are asked to calculate the ph for a solution that consists of a solution that is 0 .250 in sodium formate.

00:34 And i'll just write the solution or the chemical formula for sodium formate.

00:44 The solution is also 0 .150 molar in formic acid.

00:48 And the chemical formula for formic acid is h -c -o -o -h.

01:07 This is going to dissociate into h -c -o -o -1 -1, which is the ion we are going to use when we're doing our calculation.

01:21 Let's write our chemical equation.

01:26 Our acid will partially dissociate into our hydrogen ion plus our format ion.

01:53 If we put together an ice problem for this, we get 0 .150 and 0 .2 .0 as our initial concentrations.

02:13 Our change will be minus x, plus x, and plus x, giving me equilibrium concentrations as i'm listing here.

02:34 That's a 5.

02:35 I should make that a little nicer.

02:39 There we go.

02:44 From our appendix d, we can look up our ka, which is 1 .8 times 10 to the minus 4.

03:01 And we know our dissociation constant will equal a hydrogen ion concentration times the formate ion concentration divided by the formic acid concentration.

03:26 Let me go ahead and substitute these values.

03:29 Since our value of x is going to be very small compared to our k a and all the other concentrations i'm going to ignore those two parts and i can simplify this as follows to oh where did i write this 0 .250 x or 0 .150 this is very easy to solve for x now and for part a my x was 1 .2 .0 .0 .5.

04:42 This is very easy to solve for x now 0 .08 times 10 to the minus 4th molar.

04:48 And that's my hydrogen ion concentration.

04:52 To find the ph, we take the negative log of 1 .08 times 10 to the minus 4th molar.

05:04 And i got 3 .97 as my ph.

05:10 Our first problem of the 3 that we need to do is done.

05:16 Our next problem, p, is as follows.

05:24 Again, we're going to calculate ph.

05:26 And this time my solutions are 0 .490 molar in puridine that is c5h5n and we're 0 .450 molar in periodium chloride, periodinium chloride which is c5h5 and hcl.

06:16 Like we always do let's write a chemical equation.

06:18 I'll switch colors to keep us focused.

06:43 C5h5n plus, i forgot my water here.

06:54 Then i'm going to have my anion, it's my cation plus oh, which will also be.

07:20 Okay.

07:22 Then from here, again, we're going to do ice, and we will have 0 .490.

07:35 Don't care about that.

07:37 Here we have 0 .450 and 0 of plus x.

07:44 Plus x and minus x 0 .490 minus x 0 .450 plus x and x.

08:01 Let's check here now what is my kb for this.

08:05 My kb is 1 .7 times 10 to the minus 9 and my equation for kb will be my oh -h concentration times the concentration of the cat ion and the concentration of the base.

09:07 And again, we get to ignore our plus x and minus x...

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