Question 7 of 20 Map Sapling Learning rong base is dissolved in 645 mL of 0.200

Question

Question 7 of 20 Map Sapling Learning rong base is dissolved in 645 mL of 0.200 M weak acid (Ka 3.66 x 10 6) to make a buffer with a pH of 4.14. Assume that the volume remains constant when the base is added. OH (aq Calculate the pKa value of the acid and determine the number of moles of acid initially present. Number Number pK mol HA When the reaction is complete, what is the concentration ratio of conjugate base to acid? Number HA How many moles of strong base were initially added? Number mol OH Previous ® Give Up & View Solution Check Answer Next Exit 8 Hint

Explanation / Answer

Given:

V = 645 mL

Molarity = 0.200 M

Ka = 3.66*10^-5

pH of buffer required = 4.14

1.

Let us calculate pKa of the acid

pKa of the acid = -log(Ka) = -log(3.66 * 10) = 4.436

2.

No. of moles of the acid initially present = (0.200 mol/L) × (645/1000 L) = 0.129 mol HA

3.

Let HA be the formula of the weak acid.

After addition of strong base, a part of HA reacts with the strong base to form A ions.
Consider the dissociation of HA in the buffer solution.
HA(aq) + OH-(aq) = H2O(l) + A- (aq)

pH = pKa + log([A]/[HA])
4.14 = 4.436 + log([A]/[HA])
log([A]/[HA]) = -0.296
[A]/[HA] = 10^(-0.296) = 0.506

The concentration ratio of conjugate base to acid = 0.506 : 1

4.
Let y mole be the number of moles of the strong base initially added.

HA(aq) + OH-(aq) = H2O(l) + A- (aq)

The final concentration ratio of A to HA is
y/(645/1000) : (0.129 - y)/(645/1000) = 0.506 : 1
y : (0.129 - y) = 0.506 : 1
y × 1 = 0.506 × (0.129 - y)
y = 0.506 × 0.129 - 0.506y
1.506y = 0.506 × 0.129
y = 0.506 × 0.129 / 1.506
y = 0.043

No. of moles of strong base initially added = 0.043 mol OH-

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