clearer picture here: https://imgur.com/C5Qd9Z8.jpg trying to solve question 2 p

Question

clearer picture here: https://imgur.com/C5Qd9Z8.jpg

trying to solve question 2 part a and b.

pus curve for a monoprotie weal, o loss M Naol is added to a 25.00mLponios of acid. of base adagd to the acid. al Determine the mol l) by Determine the mole of acid present in the 25.00 ml. sample. 55 z z male. Determine the molarity of the acid. d) How many moles would he in a 10 ml sample of the acid described abone 4pts) The midpoint for the weak acid described above PH of 4.35. Determina: Ibe conditions above, at what valutte oftarami do bi Given the conditions above, al what valu de oftitran does this occur?

Explanation / Answer

2)

Dissociation of a weak monoprotic acid –

HA(aq)   <======> H3O+ + A-

Ka = [H3O+][A]/[HA]

The midpoint of a titration is the point at which exactly enough acid (or base) has been added to neutralize one-half of the acid (or the base) originally present and occurs halfway to the equivalence point.

Then Ka = [H3O+]

Taking the negative logarithm of both sides,

logKa = log [H3O+]

logKa= pKa

log [H3O+] = pH

So, pKa=pH

Thus the pH at the midpoint of the titration of a weak acid is equal to the pKa of the weak acid

logKa= pKa

Ka = 10^-PKa

Ka = 10^ -4.35 = 4.47 x 10^ -5

a)

Calculate the moles of the base-

Moles of the base = L base x M base

    = 24.00/1000 x 0.1068

    = 0.002563 moles

Moles of acid in 25ml sample-

Moles of acid = Moles of base   0.002563 moles = 2.56 x 10^-3 moles

Molarity of acid   Moles of acid/L of acid = 2.56 x 10^-3 moles /0.025L

Molarity of acid = 0.102528 M

Mb Vb = Ma Va

0.1068 x 24 = 0.1025 Va

Va = 0.1068 x 24/0.102528 = 25 ml

b) At given condition equivalence point occur when 25ml of 0.102528 M monoprotic acid is titrated with 24 ml of 1068 M NaOH.

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