Question #27dc1

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Question #27dc1

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Answer 1 · 2017-05-22T18:50:18

You add the acid to the water, and never the reverse..........

Explanation:

We want a $1 \cdot L$ $1*L$ volume of $H C l$ $HCl$ solution in water (of course I need to make these assumptions in order to answer the question).

By definition $p H = - {log}_{10} [H_{3} O^{+}]$ $pH=-log_(10)[H_3O^+]$

And thus $[H_{3} O^{+}] = 10^{- 3} \cdot m o l \cdot L^{- 1}$ $[H_3O^+]=10^-3*mol*L^-1$

And so I need to take a $1 \cdot m L$ $1*mL$ volume of $H C l$ $HCl$ , whose concentration is $1.0 \cdot m o l \cdot L^{- 1}$ $1.0*mol*L^-1$ , and add this a $1 \cdot L$ $1*L$ volumetric flask full of $0.9 \cdot L$ $0.9*L$ water, and then make the volume to $1 \cdot L$ $1*L$ .

$Concentration = \frac{Moles of solute}{Volume of solution}$ $"Concentration"="Moles of solute"/"Volume of solution"$

$= \frac{1.0 \times 10^{- 3} \cdot L \times 1 \cdot m o l \cdot L^{- 1}}{1 \cdot L} = 10^{- 3} \cdot m o l \cdot L^{- 1}$ $=(1.0xx10^-3*Lxx1*mol*L^-1)/(1*L)=10^-3*mol*L^-1$ .

$p H = - {log}_{10} (10^{- 3}) = - (- 3) = + 3$ $pH=-log_10(10^-3)=-(-3)=+3$ , as required..................

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Question #27dc1

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