If the pH is 6, what is the concentration of the hydrogen ions?
A. 10-1M
B. 10-6M
C. 10-7M
D. 10-8M
2)If the [H+] in a solution is 1×10-1mol/L, what is the [OH-]?
A. 1 × 10-1M
B. 1 ×10-15M
C. 1 ×10-13M
D. cannot be determined
3)If [OH-] > 1 x 10-7 mol/L, the solution is
A. basic.
B. acidic.
C. neutral.
4)The hydronium ion concentration for an aqueous solution in which the hydroxide ion concentration is 1 x 10-10 is
A. [H3O+] = 1 x 10-4
B. [H3O+] = 1 x 10-10
C. [H3O+] = 1 x 10-24
D. [OH-] = 1 x 10-4
5)A 25.00 mL sample of a solution of nitric acid, HNO3, is titrated with a 0.1570M solution of sodium hydroxide. The titration reaches the end point when
31.77 mL of the NaOH solution has been added. What is the molarity of the nitric acid solution?
A. 4.99M
B. 0.1570M
C. 0.3992M
D. 0.1996M
A. 10-1M
B. 10-6M
C. 10-7M
D. 10-8M
2)If the [H+] in a solution is 1×10-1mol/L, what is the [OH-]?
A. 1 × 10-1M
B. 1 ×10-15M
C. 1 ×10-13M
D. cannot be determined
3)If [OH-] > 1 x 10-7 mol/L, the solution is
A. basic.
B. acidic.
C. neutral.
4)The hydronium ion concentration for an aqueous solution in which the hydroxide ion concentration is 1 x 10-10 is
A. [H3O+] = 1 x 10-4
B. [H3O+] = 1 x 10-10
C. [H3O+] = 1 x 10-24
D. [OH-] = 1 x 10-4
5)A 25.00 mL sample of a solution of nitric acid, HNO3, is titrated with a 0.1570M solution of sodium hydroxide. The titration reaches the end point when
31.77 mL of the NaOH solution has been added. What is the molarity of the nitric acid solution?
A. 4.99M
B. 0.1570M
C. 0.3992M
D. 0.1996M
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Hi Tony P again!
1) B, 10^-6M. The equation relating pH and H+ ions is pH=-log[H+]. Thus, if the pH is 6, plug it into the equation and solve. 6=-log[H+]. [H+] = 10^-6M and that is your answer!
2) C, since [H+][OH-] = 1x10^-14. We know that the concentration of H+ is 1×10^-1 and so when we plug it back into the equation: [1×10^-1][OH-]=1x10^-4. Solve for [OH-] = 1x10^-13M.
3) A, basic. A solution where the concentration of OH- is higher than 1 x 10-7M means it has a pOH of over 7, which makes it definitely a base. The equation for pOH is very similar to pH except instead of H+ concentration, we use OH- concentration. If we solve the equation by inserting an OH- concentration higher than 1 x 10-7M, we will get a pOH of over 7 and all solutions with a pOH of over 7 are basic.
4) A, This question is very similar to number 2 in that the concentration of H+ times OH- equals 1x10^-14. Therefore, let's once again use that equation ([H+][OH-] = 1x10^-14) to solve for H+. Take note that H+ and H3O+ are used interchangebly because technically in every aqueous solution, that H+ combines with a nearby H2O to form H3O+. [H3O+][1 x 10^-10]=1x10^-14 and so [H3O+] equals 1x10^-4.
1) B, 10^-6M. The equation relating pH and H+ ions is pH=-log[H+]. Thus, if the pH is 6, plug it into the equation and solve. 6=-log[H+]. [H+] = 10^-6M and that is your answer!
2) C, since [H+][OH-] = 1x10^-14. We know that the concentration of H+ is 1×10^-1 and so when we plug it back into the equation: [1×10^-1][OH-]=1x10^-4. Solve for [OH-] = 1x10^-13M.
3) A, basic. A solution where the concentration of OH- is higher than 1 x 10-7M means it has a pOH of over 7, which makes it definitely a base. The equation for pOH is very similar to pH except instead of H+ concentration, we use OH- concentration. If we solve the equation by inserting an OH- concentration higher than 1 x 10-7M, we will get a pOH of over 7 and all solutions with a pOH of over 7 are basic.
4) A, This question is very similar to number 2 in that the concentration of H+ times OH- equals 1x10^-14. Therefore, let's once again use that equation ([H+][OH-] = 1x10^-14) to solve for H+. Take note that H+ and H3O+ are used interchangebly because technically in every aqueous solution, that H+ combines with a nearby H2O to form H3O+. [H3O+][1 x 10^-10]=1x10^-14 and so [H3O+] equals 1x10^-4.
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