How to find pH by mixing two solutions of different concentrations?


> 
> A mixture is made by combining 110 mL of 0.15 M $\ce{HCl}$ and 215 mL of 0.055 M $\ce{HI}$. What is the pH of the solution?
> 
> 
> 


$$\ce{HCl -> H+ + Cl-}$$
$$\ce{HI -> H+ + I-}$$ 
According to what I thought, the molarity of $\ce{H+}$ is the same as $\ce{HCl}$, because it is a strong acid and the mole ratio. So, I added the molarity of both acids: $\pu{0.15M} + \pu{0.055M} = \pu{0.205M}$ for $\ce{H+}$ in solution. 


$$\mathrm{pH} = -\log[\ce{H+}]$$ 
$$\mathrm{pH} = -\log[0.205]$$ 
$$\mathrm{pH} = 0.689$$




You are right in assuming the two acids are strong acids and therefore completely dissociated, but you can easily look that also up, e.g. [here](http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf).
Therefore you can assume, that the concentration of protons is equal to the concentration of the acid.
\begin{align}
c\_{\ce{HI}}(\ce{H+}) &= c(\ce{HI}) &
c\_{\ce{HCl}}(\ce{H+}) &= c(\ce{HCl}) &
\end{align}


Now you need to determine the amount of the protons in each solution with $$n=c\cdot V.$$


You have to add the amount of protons in each solution and determine the combined volume.
\begin{align}
n\_\mathrm{tot}(\ce{H+}) &= n\_{\ce{HI}}(\ce{H+}) + n\_{\ce{HCl}}(\ce{H+})&
V\_\mathrm{tot} &= V(\ce{HI}) + V(\ce{HCl})
\end{align}


To calculate the pH of the resulting concentration, just take the negative logarithm.
$$\mathrm{pH} = -\log c\_\mathrm{tot}(\ce{H+}) = -\log\left[\frac{n\_\mathrm{tot}(\ce{H+})}{V\_\mathrm{tot}}\right]$$



> 
>  Your solution should be $\mathrm{pH}=1.06$, rounded to three significant figures.
> 
> 
> 


The approach you chose can only work, if you have equal volumes of the two solutions.


 10

$$\begin{align}
M\_1V\_1+M\_2V\_2&=M\left(V\_1+V\_2\right)\\
0.015\times110+0.055\times215&=M\left(110+215\right)\\
1.65+11.825&=M\cdot325\\
M&=\frac{13.475}{325}\\
M&=0.04146=4.416\times10^{-2}\\
\mathrm{pH}&=-\log\left(4.416\times10^{-2}\right)\\
\mathrm{pH}&=-\log4.416+2\log10\\
\mathrm{pH}&=-0.617+2\\
\mathrm{pH}&=1.383
\end{align}$$


You can go this way also
$$\begin{align}
\text{HCl conc}&=0.015\ \mathrm M\\
&=0.015\times\frac{110}{1000}\ \mathrm{mol}\\
&=1.65\times10^{-3}\ \mathrm{mol}
\end{align}$$


$$\begin{align}
\text{HI conc}&=0.055\ \mathrm M\\
&=0.055\times\frac{215}{1000}\\
&=11.825\times10^{-3}\ \mathrm{mol}
\end{align}$$


$$\begin{align}
\text{Total}\ [\ce{H+}]&=\left(1.65\times10^{-3}+11.825\times10^{-3}\right)\ \mathrm{mol}\ \text{in}\ 325\ \mathrm{ml}\\
&=13.475\times10^{-3}\ \mathrm{mol}\ \text{in}\ 325\ \mathrm{ml}\\
\text{so in}\ 1000\ \mathrm{ml}&=13.475\times10^{-3}\times\frac{1000}{325}\\
&=4.146\times10^{-2}\\
\text{Thus}\ \mathrm{pH}&=1.383
\end{align}$$


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