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Titrate 10 cm3 of 0.1 M NH3 solution with 0.1 M HCl solution (factor= 0.998). After how many cm3 of HCl have been added, the pH of the oldst will be 8.8?

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"oldst"?

NH3 + HCl ==> NH4Cl

moles NH3 initially present = 10 cm3 x 1 1 L/1000 cm3 x 0.1 mol/L = 0.001 mol NH3

When adding HCl to NH3, you form a buffer solution since you have a weak base (NH3) and the conjugate acid (NH4+). Using a form of the Henderson Hasselbalch equation, we can solve for the [conj.acid]/[base] ratio as follows:

pOH = pKb + log[conj.acid]/[base] and pOH = 14 - pH = 14 - 8.8 = 5.2

pOH = 5.2 = 4.75 + log x

log x = 0.45

x = [NH4+] / [NH3] = 2.8

NH3 ===> NH4+

0.001...........0........initial

-x...............+x.........change

0.001-x.......x........equilibrium

x / 0.001-x = 2.8

x = 7.4x10-4 mole = moles NH3 reacted = moles HCl to be added

volume HCl needed: (x L)(0.1 mol/L) = 7.4x10-4 mol and x = 7.4x10-3 L = 7.4 ml

Use the ICE table to check this answer:

NH3 + HCl ===> NH4Cl

0.001.......0.00074...............0........initial

-0.00074...-0.00074....+0.00074 ... change

0.00026....0..............0.00074........equilibrium

NH4+ / NH3 = 0.00074 / 0.00026 = 2.8 = desired ratio calculated above

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