HCOOH um, the H concentration may be calculated from the pH. H 203 -log [HJi [H-

Question

HCOOH um, the H concentration may be calculated from the pH. H 203 -log [HJi [H-9.3x 10- mole/iter is a solution containing only formic acid, then [H'] [HCoo] HCOOH]-0500-0.009 0.491 mole/liter 9.3 x 10 0.491-= 1.8 x 10-4 Problems (The pH and poH Values of Aqueous Solutions) 15-12. Determine the pH of the following solutions. Assume that the electrolytes are 100% dissociated. (a) 5.2 × 10-1 M HCI (f) 6.1 × 10-5 M NaOH (b) 7.6 x 10-3 M HNO3 (g) 1.8 x 10-4 M KOH (c) 9.2 x 10- M HBr (h) 5.0 x 103 M Ba(OH)2 (d) 8.0 x 106 M H2SO (i) 6.2 x 102 M LiOH (e) 3.3 x 104 M HI 15-13. Determine the pH and pOH of each of the following solutions: (a) 0.075 M NH3, for which K, 1.8 x 10- (b) 0.50 M NH3, for which K, -1.8 x 10-5. (c) 1.00 M HC2H,02, for which K -1.8 x 10-5, 15-14. The pH of a 1.00 M solution of trichloroacetic acid, HC2CI,O2 is 0.44. What is the equilibrium constant for the acid dissociation reaction (j) 5.3 x 10-5 M NaOH

Explanation / Answer

15.12. We know that pH = -log [H+]

(a) For a monoprotic strong acid like HCl, [HCl] = [H+].
Therefore [H+] = 0.520 M and pH = -log 0.52 = 0.284.

(b) Same in case of HNO3. It is also monoprotic strong acid [HNO3] = [H+].
Therefore [H+] = 7.6X10-3 M and pH = -log 7.6X10-3 = 2.12

(c) In case of HBr. It is also monoprotic strong acid [HBr] = [H+].
Therefore [H+] = 9.2X10-5 M and pH = -log 9.2X10-5 = 4.036

(d) Same in case of H2SO4. It is diprotic strong acid [H2SO4] = 2[H+].
Therefore 2[H+] = 2(8X10-6 M) and pH = -log 16X10-6 = 4.8

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