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What is the pH of a 2.0 x 10^-4 M solution of nitric acid (HNO3)

User Subburaj
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2 Answers

1 vote

Hello!

datos:

Molarity =
2.0*10^(-4)\:M\:(mol/L)


ps: The ionization constant of the nitric acid is strong (100% ionized in water) or completely dissociates in water, so the pH will be:


pH = - log\:[H_3O^+]


pH = - log\:[2*10^(-4)]


pH = 4 - log\:2


pH = 4 - 0.30


\boxed{\boxed{pH = 3.70}}\end{array}}\qquad\checkmark

Note:. The pH <7, then we have an acidic solution.

I Hope this helps, greetings ... DexteR!

User Gary Mendonca
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6.0k points
1 vote

Answer:

The pH of the nitric acid solution is 3.70.

Step-by-step explanation:

The pH is negative logarithm of hydrogen ion concentration.


pH=-\log[H^+]

Concentration of nitric acid =
2* 10^(-4) M


HNO_3(aq)\rightarrow H^+(aq)+NO_(3)^(-)(aq)

1 mole of nitric acid gives 1 mole of hydrogen ions.

Then
2* 10^(-4) M of nitric acid will give .


[H^+]=1* 2* 10^(-4) M=2* 10^(-4) M

The pH of the solution :


pH=-\log[2* 10^(-4) M]=3.70

User Miki
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5.5k points