Calculate the pH after 0.15 mole of solid NaOH is added to 1.00 L of each of the following solut

Name: Problem 44, Chapter 15: Chemistry
Uploaded: 2020-03-25T16:38:39-08:00
Duration: 5 min 25 s
Channel: Ahmed Ali
Description: Calculate the pH after 0.15 mole of solid NaOH is added to 1.00 L of each of the following solutions: a. 0.050M propanoic acid (HC3 H5 O2, K2=1.3 ×10^-5) and 0.080M sodium propanoate b. 0.50M propanoic acid and 0.80M sodium propanoate c. Is the solution in part a still a buffer solution after the NaOH has been added? Explain.

Calculate the pH after 0.15 mole of solid NaOH is added to...

Calculate the pH after 0.15 mole of solid NaOH is added to 1.00 $\mathrm{L}$ of each of the following solutions: a. 0.050$M$ propanoic acid $\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{2}=1.3 \times 10^{-5}\right)$ and 0.080$M$ sodium propanoate b. 0.50$M$ propanoic acid and 0.80$M$ sodium propanoate c. Is the solution in part a still a buffer solution after the NaOH has been added? Explain.

Calculate the pH after 0.15 mole of solid NaOH is added to 1.00 $\mathrm{L}$ of each of the following solutions:
a. 0.050$M$ propanoic acid $\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{2}=1.3 \times 10^{-5}\right)$ and 0.080$M$ sodium propanoate
b. 0.50$M$ propanoic acid and 0.80$M$ sodium propanoate
c. Is the solution in part a still a buffer solution after the NaOH has been added? Explain.

Key Concepts

Neutralization Reaction in Buffers

When a strong base is added to a buffer, it reacts preferentially with the weak acid, converting it into its conjugate base. This reaction alters the ratio of the conjugate base to the weak acid, which can then be analyzed using the Henderson-Hasselbalch equation to determine the resulting pH change.

Buffer Capacity

Buffer capacity refers to the amount of added acid or base that a buffer can neutralize before the pH begins to change significantly. It is dependent on the absolute concentrations of the buffer components; once a substantial fraction of either the weak acid or its conjugate base is consumed, the buffer's ability to resist pH changes is diminished.

Buffer Solutions

A buffer solution consists of a weak acid and its conjugate base (or a weak base and its conjugate acid) that work together to resist large changes in pH when small amounts of strong acid or strong base are added. This is due to the equilibrium between the acid and base forms, which shifts to counteract added hydrogen or hydroxide ions.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a practical tool 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, providing a straightforward approach to understanding how the proportions of acid and base influence the pH.

Transcript

00:01 Everyone my name is ahmed ali in this question we have a mixture of propionic acid which we will write it as hbr and the sodium proponate the concentration of propionic acid is 0 .05 and the concentration of sodium proponate 0 .08 molarity and due to the volume is 1 litre so this value may be its number of more.

00:37 Similar in part b we have propionate but at this time the concentration is 0 .5 and 0 .8 and due to the volume is 1 liter so this value may be also the number of mull or the mularity because the volume is one liter.

01:01 Both mixture are puffer.

01:03 So what happened for the bh if we add here 0 .15 mole sodium hydroxide and what happened here if we add 0 .15 more sodium hydroxide.

01:21 In this case the number of mole of sodium hydroxide is higher than the number of more of acid so the buffer is destroyed because the buffer resistance appeared only if we add a small amount of strong acid or strong base.

01:44 So we what happened is sodium hydroxide react with propionic acid and we found that the amount of sodium hydroxide remain.

01:56 So the amount of sodium hydroxide remain now is 0 .1 mole because we subtracted this value from this value and the amounts that reacted are increased the amount of sodium proponate so the amount of sodium proponate is 0 .08 plus the amount of reactive between noh and hbr which is 0 .05 so the value now is 0 .08 plus 0 .05 to be 0 .13.

02:41 More.

02:43 So now we have mixture of strong base and the basic salt.

02:48 When we calculate the value of concentration of oh, we usually depend on the oh exit from strong base comparing to the basic salt.

03:02 So the concentration of oh is a 0 .1 mul per 1 liter solution.

03:13 So the concentration of oh is 0 .0 .mol, which is equivalent to 0 .1 molarity.

03:21 All of this because the volume we suppose is 1 liter...

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