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Maru [420]
3 years ago
6

How much 6.0 M HNO3 is needed to neutralize 39 mL of 2.0 M KOH?

Chemistry
2 answers:
Eva8 [605]3 years ago
7 0
<span>KOH+HNO3--->KNO3+H2O
CA=6.0M, CB=2.0M, VA=?, VB=39ml, na=1, nb=1
CAVA/CBVB=na/nb
6*VA/2*39=1/1
6VA/78=1
6VA=78
VA=78/6. VA=13ml.
VA=13ml. </span>
san4es73 [151]3 years ago
4 0

Answer : The volume of HNO_3 needed are, 0.013 L

Explanation :

According to the neutralization law,

M_1V_1=M_2V_2

where,

M_1 = molarity of HNO_3 = 6.0 M = 6.0 mole/L

M_2 = molarity of KOH = 2.0 M = 2.0 mole/L

V_1 = volume of HNO_3 = ?

V_2 = volume of KOH = 39 ml = 0.039 L

conversion : 1 L = 1000 ml

Now put all the given values in the above formula, we get the volume of HNO_3

6.0mole/L\times V_1=2.0mole/L\times 0.039L

V_1=0.013L

Therefore, the volume of HNO_3 needed are, 0.013 L

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