A 100.0-mL buffer solution is 0.175 M in HClO and 0.150 M in NaClO a. What is the initial pH of

step 1 In this question, we have 100 milliliters of a buffer solution. The buffer solution consists of

A 100.0-mL buffer solution is 0.175 M in HClO and 0.150 M in...

A 100.0-mL buffer solution is $0.175 \mathrm{M}$ in HClO and $0.150 \mathrm{M}$ in $\mathrm{NaClO}$ a. What is the initial $\mathrm{pH}$ of this solution? b. What is the $\mathrm{pH}$ after addition of $150.0 \mathrm{mg}$ of $\mathrm{HBr}$ ? c. What is the $\mathrm{pH}$ after addition of $85.0 \mathrm{mg}$ of $\mathrm{NaOH}$ ?

A 100.0-mL buffer solution is $0.175 \mathrm{M}$ in HClO and $0.150 \mathrm{M}$ in
$\mathrm{NaClO}$
a. What is the initial $\mathrm{pH}$ of this solution?
b. What is the $\mathrm{pH}$ after addition of $150.0 \mathrm{mg}$ of $\mathrm{HBr}$ ?
c. What is the $\mathrm{pH}$ after addition of $85.0 \mathrm{mg}$ of $\mathrm{NaOH}$ ?

Key Concepts

Acid–Base Buffers

Acid–base buffers consist of a weak acid and its conjugate base (or a weak base and its conjugate acid) and are used to maintain a relatively constant pH when small amounts of a strong acid or base are added. They work by neutralizing added acids or bases, thus preventing drastic changes in the hydrogen ion concentration of the solution.

Henderson–Hasselbalch Equation

The Henderson–Hasselbalch equation is a mathematical expression used to estimate the pH of a buffer solution. It relates the pH to the pKa of the weak acid and the ratio of the concentrations of the conjugate base to the acid, allowing for quick calculation of the pH under equilibrium conditions.

Reaction Stoichiometry in Buffers

Reaction stoichiometry in buffers involves understanding how the strong acid or base added to the buffer reacts with its components. The added acid will react with the conjugate base, whereas the added base will react with the weak acid. Correct stoichiometric calculations are critical for determining the resultant concentrations of the buffer components and the subsequent shift in pH.

Conversion of Mass to Moles

Conversion of mass to moles is a fundamental concept necessary to quantify the number of particles, which is particularly important when dealing with additions of reactants (like strong acids or bases) to a buffer. This conversion uses the molar mass of the substance to translate a given mass into the number of moles, which can then be employed in stoichiometric calculations to assess the effect on the system's pH.

Transcript

00:03 In this question, we have 100 milliliters of a buffer solution.

00:11 The buffer solution consists of 0 .175 molar h -c -l -o and 0 .150 molar n -a -c -l -o, or part a.

00:35 What is the initial ph? so the k .a of hcl is equal to 2 .9 times 10 to the minus 8.

00:47 The p -k -a of h -c -lo is equal to negative log of 2 .9 times 10 to the minus 8, which is equal to 7 .54.

01:01 To find the ph of the buffer, the ph, we'll use.

01:05 The henderson -hausaubach equation is equal to p -k -a plus the log conjugate base and conjugate acid which is equal to p -k -a plus the log of clo minus over h -c -l -o p -k -a is equal to 7 .54 plus the log of 0 .150 over 0 .175 and we find that the ph, the initial ph of the buffer is equal to 7 .47.

02:02 For part b, we're asked to calculate the ph after adding 150 milligrams or 0 .150 grams of hydrobromic acid.

02:18 Let's solve for the moles of hbr, 0 .150 grams of hbr.

02:27 Molar mass is 80 .9 grams in one mole, and this is equal to 0 .00185 moles of hbr.

02:45 Let's solve for the moles of h -c -l -o initial, which is equal.

02:54 To 0 .175 molar times the volume.

02:59 1 .00 liters and this is equal to 0 .0175 moles of hcl.

03:09 And we'll solve for the moles of clo of clo minus initial, which is 0 .150 molar times 0 .100 liters.

03:21 And this is equal to 0 .0150 mol.

03:25 Of clo minus.

03:30 Now we'll set up an ice table here, h plus add clo minus to h clo.

03:40 See this ice table in terms of moles.

03:46 It's 0 .0115 moles.

03:52 0 .0175 moles.

03:56 We're gonna add 0 .070 185 moles minus 0 .00185 moles and add 0 .00185 moles.

04:14 This will yield approximately 0 .01315 moles and 0 .0935 moles.

04:25 We're going to assume no change in volume upon the addition of the 150 milligrams of hbr.

04:49 Therefore, the molarity of clo minus is equal to 0 .0 .15 moles over 0 .1 .315 molars.

05:06 This works out to 0 .1315 molar.

05:09 And the molarity of hclo is equal to 0 .01935 moles over 0 .1 -0 -0 -0 liters.

05:22 This works out to 0 .1935 molar...

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