help về ph 11

L

levietdung1998

\[\begin{array}{l}
{n_{{H^ + }}} = 0,2.2.2 = 0,8\\
{n_{O{H^ - }}} = 0,3.1 = 0,3\\
\\
{H^ + } + O{H^ - } \to {H_2}O\\
\\
\to {n_{{H^ + }\left( {du} \right)}} = 0,8 - 0,3 = 0,5\\
\to \left[ {{H^ + }\left( {du} \right)} \right] = \frac{{0,5}}{{0,5}} = 0,1 \to PH = 1
\end{array}\]
 
P

phuocthuan1998

\[\begin{array}{l}
{n_{{H^ + }}} = 0,2.2.2 = 0,8\\
{n_{O{H^ - }}} = 0,3.1 = 0,3\\
\\
{H^ + } + O{H^ - } \to {H_2}O\\
\\
\to {n_{{H^ + }\left( {du} \right)}} = 0,8 - 0,3 = 0,5\\
\to \left[ {{H^ + }\left( {du} \right)} \right] = \frac{{0,5}}{{0,5}} = 0,1 \to PH = 1
\end{array}\]

ủa o,5/0,5=1 => ph=-log[1]=o mà bạn hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh/:)
 
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