How much 6.0 m hno3 is needed to neutralize 39ml of 2 m koh

13mL

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

HNO3 + KOH —> KNO3 + H2O

From the balanced equation above, we obtained the following data:

Mole ratio of the acid (nA) = 1

Mole ratio of the base (nB) = 1

Step 2:

Data obtained from the question.

This includes the following:

Molarity of the acid (Ma) = 6M

Volume of the acid (Va) =?

Volume of the base (Vb) = 39mL

Molarity of the base (Mb) = 2M

Step 3:

Determination of the volume of the acid.

Using the equation:

MaVa/MbVb = nA/nB, the volume of the acid can be obtained as follow:

MaVa/MbVb = nA/nB

6 x Va / 2 x 39 = 1/1

Cross multiply to express in linear form

6 x Va = 2 x 39

Divide both side by 6

Va = (2 x 39)/6

Va = 13mL

Therefore, the volume of the acid (HNO3) needed for the reaction is 13mL

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i believe it is 40–40–20%

explanation: