Calculate [H3O+] for pH 1.86. Steps please.

[H₃O⁺] = [H⁺] = 10^-pH = 10⁻¹°⁸⁶ = 0.0138M in [H⁺]  

Explanation:

By definition pH = -log[H⁺] => [H⁺] = 10^-pH = 10⁻¹°⁸⁶ = 0.0138M in [H⁺]

Using your calculator ... I am using a TI-30XA scientific calculator.

Depending on the calculator, the answer may show as 0.0138, or  1.38x10⁻², or 1.38E-2 (=1.38 x 10⁻²). It is the user's job to insert dimensional units into answer of interest =>  0.0138M, or  1.38 x 10⁻²M, or 1.38E-2M.

1.38E-2 which is 1.38 x 10⁻².

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