Whilst reading about pH, it is given that concentration of water is $\pu{55.5 M}$. Let water be of $\pu{1 mol}$ at $\pu{297 K}$ and $\pu{1 atm}$ pressure. Then the concentration is $$c =\frac{n}{v} = \frac{P}{RT}\\ =\frac{1}{8.314\times 297} = 0.004 $$

But it should be $\pu{55.5 mol/L}$. Where I am going wrong?

  • 1
    $\begingroup$ The ideal gas equation (pV=nRT) isn't going to help you here, you're looking at the concentration of a liquid. $\endgroup$ – NotEvans. May 13 '17 at 11:16
  • 2
    $\begingroup$ You are calculating the concentration of water vapour (a gas); your question is about liquid water. $\endgroup$ – matt_black May 13 '17 at 13:53

First, you want to work out the number of moles in a litre ($\pu{1000 mL}$) of water: \begin{align} n &= \frac{m}{M_r}\\[1ex] &= \frac{\pu{1000 g}}{\pu{18 g mol-1}}\\ &= \pu{55.5 mol}\\ \end{align}

We know that the mass of $\pu{1000 mL}$ of water is $\pu{1000 g}$, and the molecular weight of water is around $\pu{18 g mol-1}$, therefore the calculation gives us an answer of $\pu{55.5 mol}$.

This can then be plugged into the equation for concentration: \begin{align} n &= cV\\[1ex] c &= \frac{n}{V}\\ c &= \frac{\pu{55.5 mol}}{\pu{1 dm3}}\\ &= \pu{55.5 mol dm-3}\\ \end{align}

Since $\pu{1000 mL}$ of water is precisely $\pu{1 dm3}$ (and the units for concentration are in moles per decimetre), the concentration of water is also $\pu{55.5 mol dm-3}$.

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