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Calculate the pH for each of the cases in the titration of 25.0 mL of 0.200 M pyridine, C5HN(aq) with 0.200 M HBr(aq). The 𝐾b of pyridine is 1.7×10−9.   Before addition of any HBr, after addition of 12.5 mL of HBr, after addition of 18.0 mL of  HBr,
after addition of 25.0 mL of HBr HBr, after addition of 30.0 mL of HBr HBr   Some explanation of the reasoning for the setup of the first two steps would be appreciated. From where are all of the used values obtained?  (E. the value 0.190). It is otherwise useless to me. There is no 0.190 value in the given information.
   

ERROR!

Kb = [C5H5NH^+][OH^-]/[C5H5N].

 1.7×10^−9. = x^2/.190 - x.

The 0.190 should be 0.220 M as the molality of the [C5H5N], given.

Thus, the answer is wrong along with any answer based on the value obtained using this bogus value.

Kb = [C5H5NH^+][OH^-]/[C5H5N].

 1.7×10^−9. = x^2/.190 - x.

x = 1.80 × 10^-5.

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The reaction A   P is first order reaction with respect of [A]. Consider the following data.  Time (s)      [A] mol/L      0                 1.60     10                0.40     20                0.10   The half-life of this reaction is... ◯ 20 s ◯ 5 s ◯ 10 s ◯ 30 s   Please show full steps how I find the exact seconds.

Calculate the pH for each of the cases in the titration of 25.0 mL of 0.200 M pyridine, C5HN(aq) with 0.200 M HBr(aq). The 𝐾b of pyridine is 1.7×10−9.   (a). Before addition of any HBr:   (b). after addition of 12.5 mL of HBr:   (c). after addition of 18.0 mL of  HBr:
after addition of 25.0 mL of HBr HBr, after addition of 30.0 mL of HBr  The pH of the 30.0 mL volume is NOT 1.70