Question

Calculate the volume of 1.5 M nitric acid required to neutralize 30.0 mL of 7.0 M sodium

hydroxide.

HNO3 (aq) + NaOH (aq) → H2O (l) + NaNO3 (aq)

Answer 1

0.14 liters nitric acid

Step-by-step explanation:

Molarity = moles / liter

7.0 M = x / 0.03 L

= 0.21 moles NaOH

0.21 moles NaOH x ( 1 mole HNO3 / 1 mole NaOH) = 0.21 moles HNO3

Molarity = moles / liter

1.5 M = 0.21 moles / x

x = 0.21 moles / 1.5 M

x = 0.14 Liters HNO3

Answer 2

`140mL`

Step-by-step explanation:

`M_1V_1=M_2V_2`

`M_1=1.5M` molarity of nitric acid

`V_1=?` volume of nitric acid

`M_2=7.0M` molarity of sodium hydroxide

`V_2=30.0mL` volume of sodium hydroxide

`(1.5M)(V_1)=(7.0M)(30.0mL)`

`V_1 =((7.0M)(30.0mL))/(1.5M)=140mL`

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