Asked by Sharon
If 0.630 grams of HNO3 (molecular weight 63.0) are placed in 1 liter of distilled water at 25 degrees C, what will be the pH of the solution? (Assume that the volume of the solution is unchanged by the addition of the HNO3.)
A. 0.01
B. 0.1
C. 1
D. 2
E. 3
A. 0.01
B. 0.1
C. 1
D. 2
E. 3
Answers
Answered by
DrBob222
HNO3 is a strong acid; it ionizes 100%.
moles HNO3 = grams/molar mass.
M HNO3 = moles/L soln.
pH = -log(HNO3).
moles HNO3 = grams/molar mass.
M HNO3 = moles/L soln.
pH = -log(HNO3).
Answered by
Sharon
Thank you!
Answered by
slime
whats the freaking answer
Answered by
slime xd
HNO3 is a strong acid; it ionizes 100%.
moles HNO3 = grams/molar mass.
0.63/63=0.01
M HNO3 = moles/L soln.
0.01/1=0.01
pH = -log(M HNO3).
-log(0.01) = 2
moles HNO3 = grams/molar mass.
0.63/63=0.01
M HNO3 = moles/L soln.
0.01/1=0.01
pH = -log(M HNO3).
-log(0.01) = 2
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