Calculate the pH of 2.0 * 10 -7 M (CH 3 )4N + OH - ????? Is this problem below s

Question

Calculate the pH of 2.0 * 10-7 M (CH3)4N+OH- ?????

Is this problem below similar

HNO3 is a strong acid HNO3 = H+ + NO3- we can NOT neglet the autoionization of water : H2O <=> H+ + OH- for which Kw = 1.0 x 10^-14 because [H+][OH-]= 1.0 x 10^-14 [H+]= 1.0 x 10^-7 M total concentration of H+ = 1.0 x 10^-7 + 4.8 x 10^-8=1.5 x 10^.7 M pH = 6.8

or

Concentration of OH- = 2.0 × 10-7 + 1*10^-7    = 3*10^-7 M

as concentration is very less it is necessary to add OH-present in water also.

pOH = -log(OH-) = -log 3*10^-7   = 7-log 3 = 6.523

pH = 14-6.523 = 7.477

Explanation / Answer

yes, as the solubility of (CH3)4N+OH- in water is high so it can be safely assumed that it would completely dissociate in water and the process that you have taken to find the POH and then calcuating the PH is correct.

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