Hi ya'll! This problem states:
"HClO is a weak acid ( Ka=4.0×10−8 ) and so the salt NaClO acts as a weak base. What is the pH of a solution that is 0.028 M in NaClO at 25 °C?"
From my understanding, the formulas for this problem is HClO <--> H + ClO, and then the ClO is used to make NaClO with the equation ClO + Na <--> NaClO. I know we have to convert the Ka to Kb since NaClO is a weak base. But how do I go from here?
Achieve #7 Week 2-3
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Re: Achieve #7 Week 2-3
NaClO will dissociate partially in water. The equation is NaClO + H2O —> Na(OH) + HClO. Use an ice box to set up an equation for Kb. You can derive the Kb value from Kb=Kw/Ka. We are given the mol/L of NaClO so you can find the mol/l of Na(OH) (which is [OH-]). Use [OH-] to calculate pOH. Then use pOH +pH = 14 to find pH.
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Re: Achieve #7 Week 2-3
Yeah, so I would recommend writing out the problem then doing the basic ICE equations to set up Kb so you can calculate pOH. This problem is really similar to number 8 for the same week and there is also another question on here about it that has some good responses, you would just do the opposite, solving for pOH instead of pH.
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Re: Achieve #7 Week 2-3
After calculating Kb, you would use the ICE table for the hydrolysis of ClO- and calculate the concentration of OH- molecules. If Kb is smaller than 10^-4 and the x value is less than 5% of the initial concentration, it is safe to assume x is negligible compared to .028 M. After finding the concentration of OH-, you would convert to pOH and find pH by subtracting pOH from 14.
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