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

3 Answers
Apr 21, 2018

-lg(H_(Cation))

Explanation:

pH = -lg(0.043) = 1.34

pH is 1.34.

Apr 21, 2018

Approximately 1.37.

Explanation:

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

Apr 21, 2018

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

Explanation:

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....