Question

Please help me answer this question! This is due in 45 minutes.
What is the pH of a 0.02 M solution of the strong acid, HNO3 ?​

Answer 1

Answer:

☆ pH formula:

pH= $\ -log[H^{+1}]$

☆ Solution:

》$\ HNO_{3} \rightarrow{ H^{+1}+ (NO_3)^{-1}}$

》since the solution is 0.02M, the $\ H^{+1}$ concentration will be 0.02M = 2×$10^{-2}$

as it is strong acid, dissociation will be 100%

》pH= $\ -log[H^{+1}]$

》pH= $\ -log[2×10^{-2}]$

》pH= $\ -log[2]+log[10^{-2}]$

》pH= $\ -log[2]+ 2log[10]$

》pH= $\ -0.3 +2$

》pH= 1.7 (approximately)

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Answer 2

Answer:

HNO3 is a strong acid and dissociates completely or 100% .

ie., HNO3-------->H+ +NO3–.

So H+ ion concentration=0.5=5 x 10–1 N

So pH=- log(5 x 10–1)

=0.3010

~=0.3