Respuesta :

[tex]\text{Answer: 1.26 }\cdot10^{-2}molL^{-1}[/tex]

Given that

pH = 1.90

[tex]\begin{gathered} pH=-log(H^+_3O) \\ pH\text{ = 1.90} \\ \text{Take the log of both sides} \\ 1.90=-log(H^+_3O) \\ 10^{-1.9}=H^+_3O \\ H^+_3O\text{ = }1.26\cdot10^{-2}molL^{-1} \end{gathered}[/tex]

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