Question

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

Answer 1

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.

Answer 2

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