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

${(x . C_{M} . V)}_{b a s e} = {(y . C_{M} . V)}_{a c i d}$ $(x.C_M .V)_(base) = (y.C_M .V)_(acid)$

Acid-base neutralization

${(x . C_{M} . V)}_{b a s e} = {(y . C_{M} . V)}_{a c i d}$ $(x.C_M .V)_(base) = (y.C_M .V)_(acid)$

$C_{M} = M o l a r i t y$ $C_M = Molarity$

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

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

$V_{b a s e} = \frac{{(y . C_{M} . V)}_{a c i d}}{{(x . C_{M})}_{b a s e}}$ $V_(base)= ((y.C_M .V)_(acid))/((x.C_M)_(base))$

$V_{b a s e} = \frac{2 \cdot 0.17 M 3.0 mL}{1 \cdot 0.0500 M}$ $V_(base) = ( 2*0.17 " M "3.0 " mL")/(1*0.0500" M")$

$V_{b a s e} ≅ 20 . mL$ $V_(base) ~= 20. "mL"$