Question

Calculate [h ], [clo4–], and [oh–] in an aqueous solution that is 0.170 m in hclo4(aq) at 25 °c.

Answer 1

To calculate [H+], use the concentration of HClO4. [ClO4-] is half of [H+]. [OH-] can be calculated using Kw.

Step-by-step explanation:

To calculate the hydronium ion concentration ([H+]), we use the fact that HClO4 is a strong acid and completely ionizes in water. Therefore, [H+] will be equal to the concentration of HClO4, which is 0.170 M.

To calculate the perchlorate ion concentration ([ClO4-]), we need to consider that HClO4 is a monoprotic acid, meaning it only donates one proton. Therefore, [ClO4-] will be half of [H+], which is 0.085 M.

The hydroxide ion concentration ([OH-]) can be calculated using the ion product constant for water (Kw), which is equal to 1.0 x 10^-14 at 25 °C. Since Kw = [H+][OH-], we can rearrange the equation to solve for [OH-]. Plugging in the [H+] value we calculated earlier, we get [OH-] = Kw / [H+] = (1.0 x 10^-14) / (0.170) ≈ 5.88 x 10^-14 M.

Answer 2

Equation to show how it ionizes in aqueous solutions: example: HCLO4 HCLO4 + H2O---> H3O+ + CLO4- KOH Ba(OH)2 CsOH NaO ionic equation for: 2 HClO4(aq) + Mg(OH)2(s) ¨ Mg(ClO4)2(aq) +If 200 mL of 3.70 M aqueous Ba(OH)2 and 980 mL of 0.837 M aqueous HClO4 are reacted stoichiometrically according to the balanced equation, how many milliliters of 3.70 M aqueous Ba(OH)2 remCalculate [H3O ], [ClO4–], and [OH–] in an aqueous solution that is 0.170 M in HClO4(aq) at 25 °C.ain2 H2O(l)H