"each value below represents a different aqueous solution at 25 °c. classify each solution as acidic, basic, or? are the following acidic, basic or nuetral? poh=3.37 poh=11.40 [h+]=3.8x10^-4 [h+]=5.8x10^-8 [oh-]=3.5x10^-12 [oh-]=6.6x10^-3 [h+]=1.0x10^-7 poh=7 ph=5.11 ph=12.94"

1) pOH = 3.37                The solution is basic.

2) pOH = 11.40              The solution is acidic.

3) [H⁺] = 3.8 x 10⁻⁴        The solution is acidic.

4) [H⁺] = 5.8 x 10⁻⁸        The solution is basic.

5) [OH⁻] = 3.5 x 10⁻¹²    The solution is acidic.

6) [OH⁻] = 6.6 x 10⁻³     The solution is basic.

7) [H⁺] = 1.0 x 10⁻⁷         The solution is neutral.

8) pOH = 7                    The solution is neutral.

9) pH = 5.11                   The solution is acidic.

10) pH=12.94                The solution is basic.

Explanation:

pH is a sclale from 0 to 14.

1) pOH = 3.37:

∵ pH + pOH = 14.

∴ pH = 14 - pOH = 14 - 3.37 = 10.63.

∵ pH > 7.

∴ The solution is basic.

2) pOH = 11.40:

∵ pH + pOH = 14.

∴ pH = 14 - pOH = 14 - 11.40 = 2.60.

∵ pH < 7.

∴ The solution is acidic.

3) [H⁺] = 3.8 x 10⁻⁴:

∵ pH = - log[H⁺].

[H⁺] = 3.8 x 10⁻⁴ M.

∴ pH = - log(3.8 x 10⁻⁴ M) = 3.42.

∵ pH < 7.

∴ The solution is acidic.

4) [H⁺] = 5.8 x 10⁻⁸:

∵ pH = - log[H⁺].

[H⁺] = 5.8 x 10⁻⁸ M.

∴ pH = - log(5.8 x 10⁻⁸ M) = 7.236.

∵ pH > 7.

∴ The solution is basic.

5) [OH⁻] = 3.5 x 10⁻¹²:

∵ pOH = - log[OH⁻].

[OH⁻] = 3.5 x 10⁻¹² M.

∴ pOH = - log(3.5 x 10⁻¹² M) = 11.46.

∵ pH + pOH = 14.

∴ pH = 14 - pOH = 14 - 11.46 = 2.54.

∵ pH < 7.

∴ The solution is acidic.

6) [OH⁻] = 6.6 x 10⁻³:

∵ pOH = - log[OH⁻].

[OH⁻] = 6.6 x 10⁻³ M.

∴ pOH = - log(6.6 x 10⁻³ M) = 2.18.

∵ pH + pOH = 14.

∴ pH = 14 - pOH = 14 - 2.18 = 11.82.

∵ pH > 7.

∴ The solution is basic.

7) [H⁺] = 1.0 x 10⁻⁷:

∵ pH = - log[H⁺].

[H⁺] = 1.0 x 10⁻⁷ M.

∴ pH = - log(1.0 x 10⁻⁷ M) = 7.0.

∵ pH = 7.

∴ The solution is neutral.

8) pOH = 7:

∵ pH + pOH = 14.

∴ pH = 14 - pOH = 14 - 7.0 = 7.0.

∵ pH = 7.

∴ The solution is neutral.

9) pH = 5.11:

∵ pH < 7.

∴ The solution is acidic.

10) pH = 12.94:

∵ pH > 7.

∴ The solution is basic.

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !