What is the pH of a .043 M #HCl# solution?

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Answer 1 · 2018-04-21T10:49:03

#-lg(H_(Cation))#

pH = #-lg(0.043)# = 1.34

pH is 1.34.

Answer 2 · 2018-04-21T11:22:17

Approximately #1.37#.

The #"pH"# of a solution is given by:

#"pH"=-log[H^+]#

#[H^+]# is the hydrogen ion concentration in terms of molarity

The dissociation of #HCl# is:

#HCl(aq)->H^+(aq)+Cl^(-)(aq)#

So, one mole of hydrochloric acid contains one mole of hydrogen ions. So here, there are #0.43 \ "mol/L"# of hydrogen ions.

Therefore, its #"pH"# will be:

#"pH"=-log[0.043]#

#~~1.37#

Answer 3 · 2018-04-21T12:24:03

Well, I make it #1.37#...bowling another googly...

By definition, #pH=-log_10[H_3O^+]#...i.e. #"pouvoir hydrogène"# for #[HCl]# of #0.043*mol*L^-1# concentration...is given by....

#pH-=-log_10([HCl])=-log_10(0.043)=-(-1.366)=1.37#...

And we assume here, quite reasonably, that the strong acid #HCl# undergoes complete protonolysis in the water solvent according to the reaction....

#HCl(aq) + H_2O(l) rarr underbrace(H_3O^+)_"molar concentration of 0.043 mol/L" + Cl^-#

And so we take the logarithm of the initial #[HCl]# concentration DIRECTLY....