Answer:
Ka = ( About ) 5 x 10^ - 8
Explanation:
Let us first identify the dissociation equation for this weak acid,
HA ⇌ ( H+ ) + A¯
Knowing this, we can tell what the equilibrium expression is, respectively,
Ka = ( [ H+ ] [ A¯ ] ) / [ HA ]
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Now let us use the given pH 3.99 to calculate [ H+ ], knowing that pH = −log [H+],
3.99 = - log[H+],
[H+] = 10 ^ - 3.99,
[H+] = ( About ) 1 * 10^-4 M
Substitute known values into the equilibrium expression,
Ka = [( 1 x 10^ - 4 ) ( 1 x 10^ ¯4 )] / 0.199,
Ka = ( About ) 5 x 10^ - 8
.50 M KCl because 5% is the same as .05, which makes the .50M more concentrated.
Answer:
10.5 g
Explanation:
Step 1: Given data
- Molar concentration of the solution (C): 0.243 M
- Volume of solution (V): 0.580 L
Step 2: Calculate the moles of solute (n)
Molarity is equal to the moles of solute divided by the liters of solution.
M = n/V
n = M × V
n = 0.243 mol/L × 0.580 L = 0.141 mol
Step 3: Calculate the mass corresponding to 0.141 moles of KCl
The molar mass of KCl is 74.55 g/mol.
0.141 mol × 74.55 g/mol = 10.5 g