An industrial chemist studying bleaching and sterilizing...

An industrial chemist studying bleaching and sterilizing prepares several hypochlorite buffers. Find the $\mathrm{pH}$ of (a) 0.100 $\mathrm{M} \mathrm{HClO}$ and $0.100 \mathrm{M} \mathrm{NaClO} ;(\mathrm{b}) 0.100 \mathrm{M} \mathrm{HClO}$ and $0.150 \mathrm{M} \mathrm{NaClO} ;$ (c) 0.150 $\mathrm{M} \mathrm{HClO}$ and 0.100 $\mathrm{M} \mathrm{NaClO}$ (d) 1.0 $\mathrm{L}$ of the solution in part (a) after 0.0050 $\mathrm{mol}$ of NaOH has been added.

Key Concepts

Acid Dissociation Constant (Ka) and pKa

The acid dissociation constant, Ka, quantifies the strength of a weak acid by measuring its tendency to lose a proton, while the pKa, its negative logarithm, indicates the pH at which the acid is half dissociated. Understanding these concepts is essential when employing the Henderson-Hasselbalch equation to determine the pH of buffer solutions.

Stoichiometry in Buffer Reactions

Stoichiometry is vital in buffer chemistry for calculating the exact changes in the concentrations of the weak acid and its conjugate base during reactions, such as when a strong base is added. Adjusting for these changes allows for the accurate prediction of the new pH in the system.

Buffer Solutions

A buffer solution is a system that resists changes in pH when small amounts of acid or base are added. It typically consists of a weak acid and its conjugate base (or a weak base and its conjugate acid) that work together to maintain a relatively stable pH by neutralizing added acids or bases.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a crucial tool used to calculate the pH of a buffer solution. It connects the pH to the pKa of the weak acid and the ratio of the concentrations of the conjugate base to the weak acid, facilitating a straightforward calculation of the pH in buffered systems.

Transcript

00:01 Let's find the ph of these four solutions.

00:04 An important thing to remember is that naclo dissociates completely in water into n -a -plus and c -l -o -minus.

00:15 And if we have a certain concentration of n -a -c -l -o, let's say a random .5 molar, this will imply the same concentration of n -a -plus, and clo minus.

00:37 So in this first solution, since we have 0 .1 molar of naclo, this means we also have 0 .1 molar of just clo minus.

00:52 First solutions, a, b, and c, since hclo and clo are a conjugate pair, we can use the henderson -hasselbach equation to solve for the ph.

01:20 H .c .l .o.

01:23 Has a k .a.

01:24 Value of 2 .9 times 10 to the negative 8th.

01:30 P .k .a.

01:32 Equals the negative log of k .a.

01:37 So, um, hypochorus acid has a p .k .a of, let's find the 7 .538.

01:50 So for part a, the ph is equal to 7 .538.

01:55 Plus the log of the concentration of clo minus, which is .1 molar over the concentration of hclo, which is also .1 molar.

02:11 So this is equal to 7 .538 plus the log of 1, or just 7 .538.

02:26 For part b, the ph is still 7 .538 for the p .a.

02:33 Plus the log of .15 over .1, which is equal to 7 .714.

02:46 And for part c, the ph is, again, 7 .538, plus the log of .1 over .15, and this is equal to 7 .362.

03:07 Notice that when the concentrations of acid and base are equal, as in part a, then the ph is equal to the p .a.

03:18 If we have more base, then we have acid as in part b, then the ph is higher than the pkk.

03:26 And if we have more acid than we have base, as in part c, then the ph is lower than the pkk.

03:34 In part d, we're adding sodium hydroxide to the solution in part a.

03:43 So the hydroxide ions from the sodium hydroxide will react with the weak acid to form water and conjugate base.

04:05 And now we can create an icf table.

04:10 So initially, we know we start with 0 .005 moles of hydroxide.

04:17 Ions since we have 0 .005 moles of sodium hydroxide.

04:22 And now we have to figure out how many moles of hydrochloric acid and sodium chloride we begin with because we have the concentration, but we need it in moles for the table...

Please give Ace some feedback