6B.5

[Jessica Tejero 3L]

Hi, could someone help me approach this?

6B.5 Calculate the pH and pOH of each of the following aqueous
solutions of a strong acid or base: (a) 0.0146 m HNO3(aq); (b) 0.11 m
HCl(aq); (c) 0.0092 m Ba(OH)2(aq); (d) 2.00 mL of 0.175 m
KOH(aq) after dilution to 0.500 L; (e) 13.6 mg of NaOH dissolved
in 0.350 L of solution; (f) 75.0 mL of 3.5 3 1024 m HBr(aq) after
dilution to 0.500 L.

[Osvaldo SanchezF -1H]

You can start to approach this by simply knowing that the addition of pOH and pH Is equal to 14 so (pOH+pH=14). So if you find the pH of the molecule using the molarity you found in each problem you can just subtract it to 14 and get pOH or vice versa.

[Jesse Anderson-Ramirez 3I]

You can start to approach this by simply knowing that the addition of pOH and pH Is equal to 14 so (pOH+pH=14). So if you find the pH of the molecule using the molarity you found in each problem you can just subtract it to 14 and get pOH or vice versa.

How do you solve for (c)?

[Mary Shih 3J]

You can start to approach this by simply knowing that the addition of pOH and pH Is equal to 14 so (pOH+pH=14). So if you find the pH of the molecule using the molarity you found in each problem you can just subtract it to 14 and get pOH or vice versa.

How do you solve for (c)?

multiply the molarity by two because there is two OH. plug that number into log[oh] to find the POH. PH is 14 minus the number you get for the POH

[Xavier Herrera 3H]

Since these are all strong, you can assume they will be completely dissociated. So 0.0146M of HNO3 would make 0.0146M H+. It's important to remember that 0.0092M of Ba(OH)2 would make 2 * 0.0092M of OH- since 1 molecule creates 2 OH-.

[306022612]

I was wondering why for problem 6B.5 pt A, the answer is 1.84 and not 1.836 and is 12.16 and not 12.164. The number of significant figures of the decimal is 3, so why would the places after the decimal be three too, not what the book states them as.

[Freddie P-G 1H]

Hi, could someone help me approach this?

6B.5 Calculate the pH and pOH of each of the following aqueous
solutions of a strong acid or base: (a) 0.0146 m HNO3(aq); (b) 0.11 m
HCl(aq); (c) 0.0092 m Ba(OH)2(aq); (d) 2.00 mL of 0.175 m
KOH(aq) after dilution to 0.500 L; (e) 13.6 mg of NaOH dissolved
in 0.350 L of solution; (f) 75.0 mL of 3.5 3 1024 m HBr(aq) after
dilution to 0.500 L.

Does anyone know how to approach e) and f)? I have no idea how to answer them.

[charlotteweymer_2H]

For part E you can calculate the molarity using the given chemical formula, mass of the compound, and volume (M = mol / L). I am confused about part F because I am unsure how dilution affects the problem. Can anyone help with this?

[Mary Koutures 2I]

For part (e), you must divide the gram of NaOH by its molar mass to get the number of moles. Using the moles, you must plug it into the formula: M=moles/L. From there, you can plug it into the logarithmic function to get the pOH.
For part (f), you must use the formula: M1V1=M2V2. The 75.0 mL is V1, 3.5*10^-4 M is M1, and .500 L is V2. After plugging the numbers into the formula and solving, you will be left with M2, which you can plug into the logarithmic function to get the pH.