# Chemistry

posted by .

Calculate the concentrations of H3O+ and OH- ions in a 0.25 M HClO4 solution.

• Chemistry -

HClO4 is 100% ionized; therefore, the (HCO4) = 0.25M which makes the H3O^+ = 0.25. You can obtain OH^- from Kw/(H3O^).

## Similar Questions

1. ### Chemistry

When asked to name all of the ions and molecules present in a 0.1 mol L-1 solution of CH3COOH, why are there OH- ions present if the equation is CH3COOH(aq) + H2O(l) --> CH3COO-(aq) + H3O+(aq) ?
2. ### Chemistry

Calculate the end point pH, when 25 mL of 0.01 mol/L HCl solution reacts exactly with 25 mL of 0.1 mol/L NH4OH solution. NH3 Kb = 1.8 x 10^-5 This will be the pH of NH4Cl solution. NH4+ + HOH ==> NH3 + H3O^+ Ka = Kw/Kb = (NH3)(H3O^+)/(NH4^+). …
3. ### Chem I

Ok here is another one! Concentrated aqueous HClO4 has a concentration of 14.8 M. Calculate the concentrations of ALL ions present in a solution prepared by pipetting 5.00mL of concentrated HClO4 into a 1000.0-mL volumetric flask and …
4. ### chemistry

Calculate the H3O^+ and OH^- ion concentrations in a solution that has a pH of 3.72.
5. ### chemistry

Calculate [H3O ], [ClO4–], and [OH–] in an aqueous solution that is 0.130 M in HClO4(aq) at 25 °C.
6. ### Chemistry

Calculate [H3O ], [ClO4–], and [OH–] in an aqueous solution that is 0.130 M in HClO4(aq) at 25 °C. [H+]= [ClO4-]= [OH-]=
7. ### Chemistry

Calculate [H3O ], [ClO4–], and [OH–] in an aqueous solution that is 0.130 M in HClO4(aq) at 25 °C. [H+]= [ClO4-]= [OH-]=
8. ### Chemistry

when 25.0 mL of a solution containing both Fe2+ and Fe3+ ions is titrated with 23.0 mL of 0.0200 M KMnO4 (in dilute sulfuric acid). As a result, all of the Fe2+ ions are oxidized to Fe3+ ions. Next, the solution is treated with Zn …
9. ### Chem

Calculate [H3O ], [ClO4–], and [OH–] in an aqueous solution that is 0.170 M in HClO4(aq) at 25 °C.
10. ### CHEMISTRY

Ion Concentrations 1.) A solution is prepared by dissolving 5.00 g of stannic nitrate in enough water to make 250.0 mL of stock solution. A 15.0 mL aliquot (portion) of this stock solution is then removed and added to 75.0 mL of water. …

More Similar Questions