ID: C Chapter 18 &19 Values The following values will be useful for problems in

Question

ID: C Chapter 18 &19 Values The following values will be useful for problems in these Acid K,-7.2x 10 K,=4.5 × 104 (CH),N K,=1.8×10-5 K,=3.5x104 K,-2.5 x 10 K,=3.5 × 104 [Be(OhhP. K, 4.0 x 10-10 Ki - very large HBO K. 1.2 x 102 (COOH) K,i 4.2 x 107 Ka-4.8 x 101, CH,NH, NH 1.8 x 105 HNO2 CH,COOH Kv=7.4 x 105 K,=5.0 x 1030 K,=3.0 x 10-0 K,=4.0 x 10, K,-1.0× 10-5 K,-1.0×10 K, -6.0x 10 K-5.9 x 102 Ko=6.4 x 10-5 K.-5.0×104 [Co(OH)M Fe(OH Fe(OH-)] HCo, Which one of the following salts produces basic aqueous solutions? a. sodium acetate b. ammonium acetate c. lithium bromide d ammonium sulfate e. cesium Calculate the pH of a 0.36 M HOBr solution. a. 8.42 b. 6.75 c. 5.43 d. 5.03 e. 4.52 3. 4. s. Of the following questions, which ones are kinetic rather than thermodynamic concepts? . Can substances react when they are put together? II. If a reaction occurs, how fast will it occur? 11, what is the mechanism by which the reaction occurs? IV.. If substances react, what energy changes are associated with the reaction? c. d. e. I and III I and IV and IV

Explanation / Answer

3. a. sodium acetate
CH3COONa ----------> CH3COO^- (aq) + Na^+ (aq)
CH3COO^- (aq) + H2O ---------> CH3COOH(aq) + OH^- (aq)
CH3COONa is basic nature due to anionic hydrolysis.
4.
          HOBr --------> H^+ (aq) + OBr^- (aq)
   I      0.36            0          0
   C       -x             +x         +x
   E       0.36-x         +x          +x
      Ka = [H^+][OBr^-]/[HOBr]
      2.9*10^-9 = x*x/0.36-x
      2.9*10^-9 *(0.36-x) = x^2
         x = 3.23*10^-5
      [H^+] = x = 3.23*10^-5M
      PH   = -log[H^+]
           = -log3.23*10^-5
           = 4.52 >>>answer
e.4.52
5.e. >>>answer

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---