Answer: The concentration of hydrogen ions for this solution is
.
Explanation:
Given: pOH = 11.30
The relation between pH and pOH is as follows.
pH + pOH = 14
pH + 11.30 = 14
pH = 14 - 11.30
= 2.7
Also, pH is the negative logarithm of concentration of hydrogen ions.
![pH = - log [H^{+}]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5BH%5E%7B%2B%7D%5D)
Substitute the values into above formula as follows.
![pH = -log [H^{+}]\\2.7 = -log [H^{+}]\\conc. of H^{+} = 1.99 \times 10^{-3}](https://tex.z-dn.net/?f=pH%20%3D%20-log%20%5BH%5E%7B%2B%7D%5D%5C%5C2.7%20%3D%20-log%20%5BH%5E%7B%2B%7D%5D%5C%5Cconc.%20of%20H%5E%7B%2B%7D%20%3D%201.99%20%5Ctimes%2010%5E%7B-3%7D)
Thus, we can conclude that the concentration of hydrogen ions for this solution is
.
Answer:
![MM_{acid}=140.1g/mol](https://tex.z-dn.net/?f=MM_%7Bacid%7D%3D140.1g%2Fmol)
Explanation:
Hello,
In this case, since the acid is monoprotic, we can notice a 1:1 molar ratio between, therefore, for the titration at the equivalence point, we have:
![n_{acid}=n_{base} \\\\V_{acid}M_{acid}=V_{base}M_{base}\\\\n_{acid}=V_{base}M_{base}](https://tex.z-dn.net/?f=n_%7Bacid%7D%3Dn_%7Bbase%7D%20%5C%5C%5C%5CV_%7Bacid%7DM_%7Bacid%7D%3DV_%7Bbase%7DM_%7Bbase%7D%5C%5C%5C%5Cn_%7Bacid%7D%3DV_%7Bbase%7DM_%7Bbase%7D)
Thus, solving for the moles of the acid, we obtain:
![n_{acid}=0.0215L*0.250\frac{mol}{L}=5.375x10^{-3}mol](https://tex.z-dn.net/?f=n_%7Bacid%7D%3D0.0215L%2A0.250%5Cfrac%7Bmol%7D%7BL%7D%3D5.375x10%5E%7B-3%7Dmol)
Then, by using the mass of the acid, we compute its molar mass:
![MM_{acid}=\frac{0.753g}{5.375x10^{-5}mol} \\\\MM_{acid}=140.1g/mol](https://tex.z-dn.net/?f=MM_%7Bacid%7D%3D%5Cfrac%7B0.753g%7D%7B5.375x10%5E%7B-5%7Dmol%7D%20%5C%5C%5C%5CMM_%7Bacid%7D%3D140.1g%2Fmol)
Regards.
They are alike bc they both have 13 protons and neutrons