Map o sapling leaning Strong base is dissolved in 695 mL of 0.200 M weak acid (K

Question

Map o sapling leaning Strong base is dissolved in 695 mL of 0.200 M weak acid (Ka 4.64 x 105) to make a buffer with a pH of 3.99. Assume that the volume remains constant when the base is added. HA(aq) + OH . (aq) HA(aq) + OH.. (agl H,04-A-(aq) H2O) + A-lag Calculate the pKa value of the acid and determine the number of moles of acid initially present. Number Number mol HA When the reaction is complete, what is the concentration ratio of conjugate base to acid? Number How many moles of strong base were initially added? Number mol OH

Explanation / Answer

Answer:

Volume of acid = 695 ml

Molarity of acid = 0.2 M

Ka= 4.64 × 10-5

Strong base dissolved in 695 ml of above acid to make buffer of pH 3.99

Calculate pKa and initial moles of acid as follows,

pKa = -log[Ka]

pKa = -log 4.64 × 10-5

pKa= 4.33

moles of acid = concentration * volume

= 0.200 M * 695 mL

= 139 mmoles = 0.139 mol of HA present initially

Concentration of conjugate base to acid after completion of reaction

pH = pKa + log[A-/HA]

3.99 = 4.33 + log[A-/HA]

[A-]/[HA] = - Antilog 0.34

[A-]/[HA] = 0.4570

Moles of a strong base initially added

[HA] = 139 mmoles, calculated above.

[A-] / 139 = 0.4570
[A-] = 65.53 mmoles / (695 mL) = 0.0914 M

HA(Aq)+ OH-(Aq) --> H2O(l)+ A-

From the given equation we can say that one mole of base neutralizes one mole of acid so we can write

[A-] = [OH-]

We have molarity of [A-] = 0.0914M

So, [OH-] = 0.0914 × 695 ml = 63.523 mmoles of OH-

Therefore, moles of OH- initially added are 0.06352 mol

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