Question 7 The following parts of this question are independent of each other. (

Question

Question 7 The following parts of this question are independent of each other. (30 points) Suppose 2 mol of ideal gas is expanded adiabatically from a volume of 30.0 L at 298 K to a volume of 50.0 L. What is the pressure of the gas after the expansion? 7. (a) 7. (b) Suppose you have a 1.00 M solution of acetic acid (CH,COOH) and a 1.00 M solution of sodium acetate (NaCH,COO). The pK, for acetic acid is 4.75. Describe a simple method for making 1.00 L of a buffer from these solutions at a desired pH of 4.9.

Explanation / Answer

7)

A) P = nRT/V

    P= 2*0.0821*298/30

P= 1.63atm --------------------initial pressure

P1V1= P2V2

P2 = P1V1/V2

P2 = 1.63*30/50

P2= 0.978atm -------------------------final pressure

7) b)

P^H = P^Ka + log[CH3COONa]/[CH3COOH]

    p^H = 4.75 + log[1/1]

P^H = 4.75

-log[H^+] = 4.75

   [H^+] = 10^-4.75

    [H^+] = 1.778*10^-5

P^H = 4.9

-log[H^+] = 4.9

[H^+] = 10^-4.9

[H^+] = 1.258*10^-5M

WE NEED TO DILUTE THE 1.778*10^-5M ACETIC ACID SOLUTION TO 1.258*10^-5 M

M1V1= M2V2

1.778*10^-5*V1 = 1.258*10^-5*1

V1= 0.707 LITRES

THEREFORE

W NEED 0.707L OF 1.778 *10^-5 M ACETIC ACID TO MAKE THE P^H OF THE BUFFER SOLUTION 4.9

THERE FORE

TAKEN 0.707L OF 1.778 *10^-5 M ACETIC ACID AND ADDED0.293 L OF WATER WE GET THE BUFFER SOLUTION WITH P^H 4.9

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

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