Can you help me with 9 and 14, Please. Find the pH. What are the plh va solution

Question

Can you help me with 9 and 14, Please.

Find the pH. What are the plh va solutions? (a) 0.1 M HCI (b) 0.1 M NaOH c) 0.05 M HCI (d) 0.05 M NaOH tions of H,PU, (c) pH 8.0 19, Neutralization of acid (H,PO,) can give values, sketch a plot of sodium hydroxide sol phosphoric acid at pH 7. A weak acid. What is the pH of a 0.1 M solution of acetic 20. Buffe capacity. T prepared, one with a c with a concentration o when the following con to each of these soluti acid (Hint: Let x be the concentration of Ht ions released from acetic acid when it dissociates. The solutions to a quadratic equation of the form ax + bx + c of the form ax2 + bx + c = 0 are cl S. Substituent effects. What is the pH of a 0.1 M solution of chloroacetic acid (CICH, COOH, pKa = 2.86)? and 0.05 M 21) Buffer preparation ution of ting of acetic acid an acid plus acetate conce What concentrations should you use? Assum Water in water. Given a density of 1 g/ml and a molecu weight of 18 g/mol, calculate the concentration of water n water 10. Basic fact. What is the pH of a 0.1 M solution of ethylamine, given that the pK of ethylammonium ion (CH, CH2 NH,) is 10.70? buffer, how many mol will you need? How n (molecular weights: ac acetate, 82.03 g mol-1) 11. Comparison. A solution is prepared by adding 0.01 M acetic acid and 0.01 M ethylamine to water and adjusting the pH to 7.4. What is the ratio of acetate to acetic acid? What is the ratio of ethylamine to ethylammonium ion? 22. An alternative a pp 12. Concentrate. Acetic acid is added to water until the pH value reaches 4.0. What is the total concentration of the added acetic acid? buffer described in Pr laboratory is out of sodi hydroxide. How much sodium hydroxide do 23. Another alternative 13 Dilution. 100 mL of a solution of hydrochloric acid with pH 5.0 is diluted to 1 L. What is the pH of the diluted solution? tory was out of acetic Problem 21 by dissolv water, carefully adding more water to reac total h a to concentration of a tion? Will this solutior with the desired buffer? buffer solution made from acetic acid and sodium acetate with pH 5.0 is diluted to 1 L. What is the pH of the diluted solution?

Explanation / Answer

9) Here, we need to find out the concentration. We will find it in terms of Molarity. Molarity is mole of solute over Liter of solution. In this case, we have no solute, we only have solvent water, so we make assumptions

Let's consider Volume = 1000 mL

We are given Density = 1g/ mL,

So for 1000 mL, our mass will be 1000g

Now,

mole=mass/molecular wt
mole=1000g/ 18g per mL
mole=55.56

Molarity=mol/L solution
M=55.56/1L
M=55.56M

14) We know, acid dissociation constant of acetic acid is Ka=1.76*105

So, pKa=log(Ka) = 4.754

pH= 5

pH= -log[H+]

So, [H+] = 1*10-5

We are given 0.1mM buffer solution = 0.0001 M

100 mL solution is diluted to 1 L solution.

Initially Before Dilution
Using Henderson-Hasselbalch equation equation:
pH = pKa + log([conjugate base]/[weak acid])

[conjugate base] = x

[weak acid] = 0.1 - x (since buffer is 0.1 mM)

So, 5= 4.754 + log [x/(0.1-x)]

[x/(0.1-x)] = 1.761

Solving, we get

x = 0.0637 mM

0.1-x = 0.0363 mM

[conjugate base], [A-] = 0.0000637 M

[weak acid], [HA] = 0.0000363 M

[H+] = 0.00001

Now Dilute to 1/10 concentration
[HA] = 0.00000363
[A-] = 0.00000637
[H+] = 0.000001

Note that the pH is now 6
Q = [A-][H+]/[HA] = 0.000001754 which is less than Ka = 10-4.75 =0.000017783

So equilibrium must shift away from HA and towards A- and H+

Now setting up simple equilibrium ICE table problem
HA <=> A- + H+

( 0.000001 +x)*(0.00000637 + x)/(0.00000363 - x)=10-4.75
Solve for x, x=2.1106*10-6
0.000001 + 2.1106*10-6 = 0.000003111
-log(0.00003111) = 5.51 (approx due to rounding up in the numerical)

so, pH is 5.51.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.