What volume of 0.0500M KOH solution is required to completely neutralize 3.0 mL of 0.17M H2SO4?

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Answer 1 · 2016-05-11T20:57:34

$(x.C_M .V)_(base) = (y.C_M .V)_(acid)$

Acid-base neutralization

$(x.C_M .V)_(base) = (y.C_M .V)_(acid)$

$C_M = Molarity$

$x= "number of mol ions " OH^(-) "in the formula."$

$y ="number of mol ions " H^+"in the formula."$

$V_(base)= ((y.C_M .V)_(acid))/((x.C_M)_(base))$

$V_(base) = ( 2*0.17 " M "3.0 " mL")/(1*0.0500" M")$

$V_(base) ~= 20. "mL"$