please i need a step by step solution.......thanks Strong base is dissolved in 5

Question

please i need a step by step solution.......thanks

Strong base is dissolved in 525 mL of 0.400 M weak acid (K_a = 4.38 times 10^-5) to make a buffer with a pH of 4.16. Assume that the volume remains constant when the base is added. HA(aq) + OH^- (aq) rightarrow H_2O(l) + A^-(aq) Calculate the pK_a value of the acid and determine the number of moles of acid initially present. pK_a = When the reaction is complete, what is the concentration ratio of conjugate base to acid? [A^-]/[HA] = How many moles of strong base were initially added?

Explanation / Answer

1. pKa of the acid can be calculated as follows:

pKa = -log[Ka]

pKa = -log (4.38 x 10-5)

pKa = 4.36

Therefore, pKa of given acid is 4.36

The number of moles of acid present in 525mL of 0.400M solution can be calculated as follows:

Moles of acid= Volume x molarity

Moles of acid = 0.525 L x 0.400 mol/L

Moles of acid = 0.21 moles

Thus, number of moles of weak acid present is 0.21moles

2. Concentration ratio of conjugate base to acid can be calculated using Henderson Hasselbalch equation as follows:

pH = pKa + log ([A-]/[HA])

4.16 = 4.36 + log ([A-]/[HA])

log ([A-]/[HA]) = 4.16 – 4.36

log ([A-]/[HA]) = -0.2

[A-]/[HA] = 0.631

Therefore, ratio of concentration of conjugate base to acid is 0.631

3. Let x be the moles of base initially added.

Then, Amount of acid remaining = 0.21 – x

Therefore, using expression obtained in part 2,

[A-]/[HA] = 0.631

(x/525mL) / (0.21-x)/525mL = 0.631............(525mL is the total volume)

x / (0.21-x) = 0.631

x = 0.631 (0.21-x)

x = 0.132 – 0.631x

x + 0.631x = 0.132

1.631 x = 0.132

x = 0.081

Thus, number of moles of strong base initially added is 0.081 mole

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