2HCI + Na2SO4 -> 2 NaCI + H2SO4
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<span>Data:
pH = 5.2
[H+] = ?
Knowing that: (</span><span>Equation to find the pH of a solution)</span>
![pH = -log[H+]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH%2B%5D)
<span>
Solving:
</span>
![5.2 = - log [H+]](https://tex.z-dn.net/?f=5.2%20%3D%20-%20log%20%5BH%2B%5D)
Knowing that the exponential is the opposite operation of the logarithm, then we have:
![[H+] = 10^{-5.2}](https://tex.z-dn.net/?f=%5BH%2B%5D%20%3D%2010%5E%7B-5.2%7D)
![\boxed{\boxed{[H+] = 6.30*10^{-6}}}\end{array}}\qquad\quad\checkmark](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7B%5BH%2B%5D%20%3D%206.30%2A10%5E%7B-6%7D%7D%7D%5Cend%7Barray%7D%7D%5Cqquad%5Cquad%5Ccheckmark)
pH = 5.2
[H+] = ?
Knowing that: (</span><span>Equation to find the pH of a solution)</span>
<span>
Solving:
</span>
Knowing that the exponential is the opposite operation of the logarithm, then we have: