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Question: We have \( 5g \) of \( HBr \) in \( 100ml. \) Find \( pH \) ....

We have 5g5g of HBrHBr in 100ml.100ml. Find pHpH .

Explanation

Solution

We must recall the basic definition of molarity and we must also keep in mind the correlation of volume and density of a solution. Molarity is the number of moles of solute per litre of solution.

Complete answer:
Molar concentration and pHpH can be used to convert between the mass or moles of solute and the volume of the solution. Molarity is a measurement of the moles in the total volume of the solution, whereas Molality is a measurement of the moles in relationship to the mass of the solvent. Hence, the two should never be confused.
Let us first assume that there is 100ml100ml of the resulting solution. So, we can write the given amount as 5g5g , We know, The solution contains 5g5g of HBrHBr which means the mass of HBrHBr in the solution is:
5g (HBr)=?mole 5g~\left( HBr \right)=?mole~ , molecular weight H=1, Br=79.904H=1,\text{ }Br=79.904 ,
total molecular mass =80.904=80.904 ; total molecular mass =80.904=80.904 ,
we know that, mass=mole×molarmassmass=mole\times molar mass
here, mole=5g80.904= 0.062\Rightarrow mole=\dfrac{5g}{80.904}=~0.062
there are 0.0620.062 moles in 100 ml100\text{ }ml along with 0.1 L=0.62 M0.1\text{ }L=0.62\text{ }M .
In HBrHBr , there are 0.62 M0.62\text{ }M of H+ {{H}^{+}}~ and 0.62 M0.62\text{ }M of BrBr
Thus, pHpH is given by; pH=log[H+]=log[0.62]=0.2pH=-log\left[ {{H}^{+}} \right]=-log\left[ 0.62 \right]=0.2

Note:
We should use Molality instead of Molarity in experiments involving significant temperature changes. This is because as the volume of a solution increases with temperature, heating causes molarity to decrease; however, since molality is based on mass rather than volume, molality remains unchanged.