The concentration of hydrogen ion in a sample | vedclass

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(A) Given:$\left[ H^{+} ight] = 3.8 	imes 10^{-3} \ M$The formula for $pH$ is:$pH = -\log \left[ H^{+} ight]$Substituting the value:$pH = -\log \

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$2$

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(A) Given:$\left[ H^{+} ight] = 3.8 	imes 10^{-3} \ M$The formula for $pH$ is:$pH = -\log \left[ H^{+} ight]$Substituting the value:$pH = -\log \left( 3.8 	imes 10^{-3} ight)$$pH = -(\log 3.8 + \log 10^{-3})$$pH = -(\log 3.8 - 3)$$pH = 3 - \log 3.8$Using $\log 3.8 \approx 0.58$:$pH = 3 - 0.58 = 2.42$Thus,the $pH$ of the soft drink is $2.42$.

The concentration of hydrogen ion in a sample of soft drink is $3.8 \times 10^{-3} \ M$. What is its $pH$ (in $.42$)?

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