Given that Ka = 4.16*10^-5 pKa = - log Ka = - log 4.16*10^-5 = 4.38 No. of moles

Question

Given that Ka = 4.16*10^-5

pKa = - log Ka

= - log 4.16*10^-5

= 4.38

No. of moles of the acid initially present = molarity * volume in L

= (0.200 mol/L) × (635/1000 L)

= 0.127 mol of HA acid


given that 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) + HO(l) A(aq) + HO(aq)

pH = pKa + log([A]/[HA])
4.14 = 4.38 + log([A]/[HA])
log([A]/[HA]) = -0.24

[A]/[HA] = 100.24 = 0.575

[A]/[HA] = 0.575



therefore, The concentration ratio of conjugate base to acid = 0.575 : 1

Let x mol be the number of moles of the strong base initially added.

HA(aq) + OH(aq) A(aq) + HO(l)

The final concentration ratio of A to HA is

x/(635/1000) : (0.127 - x)/(635/1000) = 0.575 : 1

x : (0.127 - x) = 0.575 : 1
x × 1 = 0.575 × (0.127 - x)

y = 0.575 × 0.127 - 0.575x

1.575 x = 0.575 × 0.127
x = 0.575 × 0.127 / 1.575
x = 0.0464

Explanation / Answer

Sapling Learning Strong base is dissolved in 635 mL of 0.200 M weak acid (Ka = 4.16 × 10-5) to make a buffer with a pH of 4.14. Assume that the volume remains constant when the base is added HA(aq) +011-(aq) -> 11,0(1) + A-(aq) 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 [HA] How many moles of strong base were initially added? Number Mig You v Click mol OH VI

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