For each strong acid solutions, determine [H3O+],[OH?], and pH. Part A 0.17 M HC

Question

For each strong acid solutions, determine [H3O+],[OH?], and pH.

Part A

0.17 M HCl

Enter your answers numerically separated by a comma. Express your answer using two significant figures.

[H3O+],[OH-] = M

Part B

Express your answer to two decimal places.

pH =

Part C

2.7×10?2 M HNO3

Enter your answers numerically separated by a comma. Express your answer using two significant figures.

[H3O+],[OH-] = M

Part D

Express your answer to two decimal places.

pH=

Part E

a solution that is 7.5×10?2 M in HBr and 1.9×10?2 M in HNO3

Enter your answers numerically separated by a comma. Express your answer using two significant figures.

[H3O+],[OH-]= M

Part F

Express your answer to two decimal places.

pH=

Part G

a solution that is 0.775% HNO3 by mass (Assume a density of 1.01 g/mL for the solution.)

Enter your answers numerically separated by a comma. Express your answer using three significant figures.

[H3O+],[OH-]= M

Part H

Express your answer to three decimal places.

Explanation / Answer

Part A and B:
HCl is a strong acid, so it will dissociate completely: HCl ------> H+ + Cl-
[H+] = 0.17 M
pH = -log(0.17) = 0.77
[OH-] = 1x10-14 / 0.17 = 5.88x10-14 M

Part C and D:
HNO3, is a strong acid too so: HNO3 -------> H+ + NO3-
[H+] = 2.7x10-2 M
[OH-] = 1x10-14 / 2.7x10-2 = 3.7x10-13 M
pH = -log(2.7x10-2) = 1.57

Part G and H:
0.775% in mass and a density of 1.01 g/mL, assuming we have 1 L of solution, let's calculate the molarity of this solution (MW = 63 g/mol)
M =0.775 * 1.01 * 1000 / 63 * 100 = 0.1242 M
[H+] = 0.1242 M
[OH-] = 1x10-14/0.1242 = 8.05x10-14 M
pH = -log(0.1242) = 0.91

PArt E and F solve it in a similar way changing values (both of them are strong acids, and the H+ concentration are additive).

Hope this helps

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