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21 October, 15:37

How many mL of 0.1 M NaOH would be required to neutralize 2.0 L of 0.050 M HCl

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  1. 21 October, 19:14
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    By using the formula M1V1=M2V2 we can calculate this

    M1=0.1 M

    M2=0.050

    V1=?

    V2=2 L

    So,

    0.1*V1=0.050*2

    V1=1 L=1000ml
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