A 75.0 mL volume of 0.200 M NH3 (Kb=1.8*10^-5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 25.0 mL of HNO3.

Question

anonymous · Answer

anybody..please help me..urgent !

anonymous · Answer

try..

rogue · Answer

$$NH_3 + HNO_3 \rightarrow NH_4 NO_3$$$$n = MV$$Moles of NH3:$$n = 0.075 L * 0.200 M = 0.015 mol NH_3$$Moles of HNO3:$$n = 0.025 L * 0.500 M = 0.0125 mol HNO_3$$

rogue · Answer

HNO3 is a strong acid, so it will dissociate completely into protons & nitrate ions. The 0.0125 mol HNO3 will neutralize 0.0125 mol NH3 and form ammonium nitrate.
0.015 mol - 0.0125 mol = 0.0025 mol NH3 remaining

rogue · Answer

$$M = \frac {n}{V} = \frac {0.0025 mol NH_3}{0.100 L} = 0.025 Molar NH_3$$Now we can make an ice chart for this remaining amount of weak base.$$NH_3 + H_2O \rightarrow NH _{4}^{+} + OH^{-}$$[Initial] 0.025 M________ 0 M_______ 0M
[Change]   -x                 +x              +x
[Final] 0.025 M - x          x                 x

rogue · Answer

$$k_b = \frac {[OH^{-}][NH4_{4}^{+}]}{[NH3]}$$$$1.8 \times 10^{-5} = \frac {x^2}{0.025 M - x}$$$$x \approx 6.62 \times 10^{-4} M$$

rogue · Answer

x is the concentration of hydroxide ions, so we can get the pOH from that.$$pOH = - \log [OH]$$$$pOH = - \log_{10} 6.62 \times 10^{-4} \approx 3.18 $$$$pOH + pH = 14$$$$3.18 + pH = 14$$$$pH = 10.82$$

rogue · Answer

That was long! Hopefully I didn't make any mistakes because there were a lot of steps...