A 0.050 M solution of the salt NaA has a pH of 9.00. Determine the [OH] and the pH of a 0.010 M solution of the acid HA. Show your work in the question parts below. Determine the equilibrium concentration of H3O* using the acid ionization equation. Recall that [OH¯] = 1.0 x 10-5. Consider the ionization equation described below.. HA(aq) + H20(1) = H;O*(aq) + A¯(aq) 1 Based on the given information, set up the ICE table in order to determine the unknown equilibrium concentration of all reactants and products. HA(aq) H20(1) H;O*(aq) A"(aq) + + Initial () Change () Equilibrium ()

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A 0.050 M solution of the salt NaA has a pH of 9.00. Determine the [OH]
and the pH of a 0.010 M solution of the acid HA. Show your work in the
question parts below.
Determine the equilibrium concentration of H3O* using the acid ionization
equation. Recall that [OH] = 1.0 x 10-5. Consider the ionization equation
described below..
HA(aq) + H20(1) =H;O*(aq) + A (aq)
1
Based on the given information, set up the ICE table in order to determine the unknown
equilibrium concentration of all reactants and products.
НА(ag)
H20(1)
H;O*(aq)
А (ag)
+
Initial ()
Change ()
Equilibrium ()
5 RESET
0.010
0.050
+x
-2х
+2x
-x
0.010 + x
0.010 - x
0.050 + x
0.050 - x
0.010 + 2x
0.010 - 2x
0.050 + 2x
0.050 - 2x
Transcribed Image Text:A 0.050 M solution of the salt NaA has a pH of 9.00. Determine the [OH] and the pH of a 0.010 M solution of the acid HA. Show your work in the question parts below. Determine the equilibrium concentration of H3O* using the acid ionization equation. Recall that [OH] = 1.0 x 10-5. Consider the ionization equation described below.. HA(aq) + H20(1) =H;O*(aq) + A (aq) 1 Based on the given information, set up the ICE table in order to determine the unknown equilibrium concentration of all reactants and products. НА(ag) H20(1) H;O*(aq) А (ag) + Initial () Change () Equilibrium () 5 RESET 0.010 0.050 +x -2х +2x -x 0.010 + x 0.010 - x 0.050 + x 0.050 - x 0.010 + 2x 0.010 - 2x 0.050 + 2x 0.050 - 2x
Expert Solution
Step 1

The pH of NaA solution = 9

The concentration of NaA = 0.050 M

pOH = 14-9

         = 5

pOH = - log (OH-)5 =- log (OH-) (OH-) =10-5 M

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