Calculate the pH of the following solution pH = 2 #HCN# & pH = 12 #NaOH# (#K_a = 10^(-10)#)?

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Answer 1 · 2018-03-22T17:08:44

see a solution process below;

#pH=pK_a +- Log ([[base]]/[[acid]])#

#pH# of #NaOH=12#

#pH+pOH=14#

#12+pOH=14#

#pOH=14-12#

#pOH=2#

#pOH=-Log[OH^-]#

#2=-Log[OH^-]#

#[OH^-]=10^-2#

#[Base]=0.01M#

#pH# for #HCN=2#

#pH=-Log[H^+]#

#2=-Log[H^+]#

#[H^+]=10^-2#

#[Acid]=0.01M#

#pK_a=-Log[K_a]#

#pK_a=-Log[10^-10]#

#pK_a=10M#

Recall;

#pH=pK_a +- Log([[base]]/[[acid]])#

#pH=10 +- Log([[0.01]]/(p[0.01]])#

#pH=10 +- Log 1#

#pH=10 +- 0#

#pH=10#