Question

Please type all answers and show all work.

You titrate 25.0mL of 0.10M NH3 with 0.10M HCl. (Kb=0.000018)

a.) What is the pH of the NH3 solution before the titration begins?

b.)what is the pH at the equivalence point?

c.)calculate the pH of the solution after adding 5.00,15.00,20.00,22.00, and 30.00mL of the acid

d.)using this info, plot the titration curve (pH vs titrant volume)

  

30.00.

Answer 1

NH_3 + H_2O NH^+ _4 + OH^- K^-_b = 1.8 times 10^-5 = [NH^+ _4] [OH^-]/[NH_3] (a) p^H before the titration: = > k_b = x^2/0.1 - x 1.8 times 10^-5 = x^2/0.1 - x = > x = [^-OH] = 0.0013 M = > p^OH = 2.89 = > p^H = 11.1 (b) p^H at equivalence point: Moles of NH_3 = 0.10 times 0.025 = 0.0025 = > moles of HCl reg = 0.0025 = > Volume of HCl required = moles of HCl/conc of HCl = 0.0025/0.1 = 0.025 Total volume (V_T) = 0.050 L \r\n Now at equivalence point, because whole of NH_3 has take part in reaction with HCl, we have [NH^+ _4] = 0.0025/0.050 = 0.05 M Also, we know that k_a times k_b = k_w = > k_a = 10^-14/1.8 times 10^-5 = 5.56 times 10^-10 and, NH^+ _4 H^+ + NH_3 = > k_a = x^2/0.05 -x = > x [H^+] = 5.3 times 10^-6 m \r\n
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the above plot is between ph and vol(in mL) of HCl added in NH3.