28. Select the solution below that is the most acidic [HyO\')-1.0× 10-4M C) (HO]

Question

28. Select the solution below that is the most acidic [HyO')-1.0× 10-4M C) (HO]-1.0x 10-M D) [H3O+]-1.0×10-s M E) [HO+]-1.0× 10-10M 29. Select the solution below that is the most basic. A) [OH-] = 1.0 x 10-10M B [OH1-1.0x 10*M [HsO']-1.0 x 106M D) [HO] 1.0x 10-7M E) [HO']-1.0 x 10- M 30. Given an OH? concentration of 1.0 10-6 M, calculate the HO. concentration, then identify the solution as acidic, basic, or neutral. A) B) C) [H30.1-1.0 × 10-7, neutral [H30.] = 1.0 x 10-6, acidic [H3O+] = 1.0 × 10-s, acidic [H3O+] = 1.0 × 10-8, basic [H:O']-1.0x 10-6, basic 31. Given an HsO" concentration of 1.0x 10* M, calculate the OH concentration, then identify the solution as acidic, basic, or neutral. A) [OH]-1.0 x 10-7, neutral B) [OH-1-1.0 x 10-4, acidic [OH]-1.0 x 10-10, acidic D) [OH]-1.0 x 10-10, basic E) [OH] = 1.0 × 10-4, basic 32. Calculate the pH of a solution that has [OH]-5.9 x 10-s M A) pH 5.90 B) pH = 5.00 ) pH=4.23 pH = 9.77 E) pH-5.59

Explanation / Answer

28. The solution which is most acidic will have the lowest pH. We are given with the concentration of H+ ions, and pH=-log[H+]. Your answer is correct as that will have a pH of 4 which is the lowest.

29. The solution which is the most basic will have highest pH. And, pH=14 - pOH and pOH=-log[OH-]. The highlighted answer is correct as that will have pH of 10, which is highest (pOH=4, therefore pH=14-4=10).

30. [OH-] = 1.0 x 10-6 M, this implies pOH=-log[1.0 x 10-6]=6. And since pH=14-pOH, so pH=14-6=8. Now, pH=-log[H+]. So, [H+]=antilog(-8)=1.0 x 10-8 M. Since pH>7, it is a basic solution.

31. Similar to question 30, this time we are given with concetration of H+ ions. So, pH=-log[H+]. pH=4, this implies pOH=14-4=10. So, pOH=-log[OH-]. Therefore, [OH-]=antilog(-10)=1.0 x 10-10. Since pH<7, it is an acidic solution.

32.  [OH-]=5.9 x 10-5M, pOH=-log(5.9 x 10-5)=-log(5.9)+5=4.22. Now, pH=14-4.22=9.77

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