the reaction and ice table:
HClO + H2O <--> H3O++ ClO-
.256M.............0........0
-x.................+x......+x
.256M -x.........+x......+x
so the expression for the reaction is
ka=[x]2/(.256M-x)
you can neglect the x in the denominator because it is very small and solve
sqrt(.256M*2.9E-8)=8.62E-5=[H3O+]
Determine pH, pOH, [H3O+] and [OH-] of a 0.265 M HClO solution. Ka of HClO is 2.9 x 10-8.
Determine the PH and [h3o+] in a .265 M HClO solution. The Ka of HClO is 2.9x10^-8
2) calculate the h3o+ in a 0.345 m hclo solution. ka = 2.9 x 10-8 find the percent ionization.
Q1)Determine the [H3O+] in a 0.256 M HClO solution. The Ka of HClO is 2.8 10–9. Q2)Find the percent ionization of a 0.373 M HF solution. The Ka for HF is 3.4 10–5. thanks!
determine the pH of a solution that is 0.15 M HClO2 (Ka=1.1 x 10^-2) and 0.15 M HClO (Ka=2.9 x 10^-8).
Drinking water is often disinfected with Cl2, which hydrolyzes to form HClO, a weak acid but powerful disinfectant: Cl2(aq) + 2 H2O(l) HClO(aq) + H3O+(aq) + Cl−(aq) The fraction of HClO in solution is defined as [HClO] / [HClO] + [ClO−] (a) What is the fraction of HClO at pH 7.00 (Ka of HClO = 2.9×10−8)? (b) What is the fraction at pH 8.50?
Determine [H3O+][H3O+] of a 0.200 MM solution of formic acid (Ka=1.8×10−4Ka=1.8×10−4).
Determine the [H3O+][H3O+] of a 0.150 MM solution of formic acid (Ka=1.8×10−4Ka=1.8×10−4).
2. For the following reactions, determine the equilibrium constants and whether the reactants or products are favored: a) HCOOH(aq) + H2O(l) → HCOO− (aq) + H3O + (aq) (To determine K, use [HCOOH] = 0.10 M and [HCOO− ] = [H3O + ] = 0.0042 M) b) H2CO3(aq) + ClO− (aq) → HCO3 − (aq) + HClO(aq) (To determine K, use Ka(H2CO3) = 4.3 × 10-7 and Ka(HClO) = 3.5 × 10-8 ) 3. For each of the above reactions...
For each strong base solution, determine [OH−], [H3O+], pH, and pOH. 1.0×10−4 M Ca(OH)2, determine [OH−] and [H3O+]. For this solution determine pH and pOH. 2.9×10−4 M KOH, determine [OH−] and [H3O+]. For this solution determine pH and pOH.