# ph of .26M methylammonium nitrate?

I have problems with a homework problem:

Calculate the pH of a 0.26 M solution of methylammonium nitrate  (CH3NH3)NO3

I believe the strong acid is HNO3 and the weak base is CH3NH3

so BA + H20 <> BOH + H+ + A-

help, how do I complete this thank you in advance for your gracious time

pH = -log [H+]

Given, [H+] = 0.26 M

Therefore, pH= - log[0.26]

=-log[26 /100]

= - log [26] + log [100]           {Since log [a/b] = log a -  log b}

=- log [26] +2

=2- log [26]

=2- 1.415                                  {Since, log 26 base 10= 1.415}

=0.583

THE ANSWER ABOVE IS INCORRECT!

The methylammoniam ion would act as a weak acid due to hydrolysis.  You would need to determine the Ka for the methylammonium ion.  This might be determined by looking up the Kb for methylammonia. (remember Ka = Kw/Kb).

Once you know the Ka you would utilize a Ka expression to determine the [H+] and from there the pH of the solution.

However, it is NOT CORRECT to assume that the [methylammonium] = [H+]  (which is what was done when it was stated   "Given, [H+] = 0.26 M"