How much (mL) of a 2.0 M sodium hydroxide solution would it take to neutralize 50 mL of a 6.0 M solution of hydrochloric acid?

150ml

Explanation:

For this question,

NaOH completely dissociates. It is a strong base

HCl also completely dissociates. It is a strong acid

So we have this equation

m1v1 = m2v2 > equation 1

M2 = 2m

V1= ??

M2 = 6m

V2 = 50m

When we input these into equation 1, we have:

2m x v1 = 6m x 50ml

V1 = 6m x 50ml/2

V1 = 300/2

V1 = 150ml

Therefore NaOH that is required to neutralize the solution of hydrochloric acid is 150ml.

Thank you

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.

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