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Calculate the pH of a solution that is 0.111 M HCl. PH=?

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HCl(aq) ------> H⁺(aq) + Cl(aq)
↓ ↓ ↓
i : 0.111 M 0 0
v: - 0.111 M +0.111 M +0.111 M
_______________________________
eq: 0 +0.111 M +0.111 M

HCl is a strong acid, so it ionizes 100% in solution :

PH = - log [ H+]

PH = - log [ 0.111]

PH = 0.9546

hope this helps!
User Paul Wray
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