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Solution and answer to problem
|
How many mL of 0.20 M HBr is needed to
neutralize 23.45 mL of 0.11 M KOH? What is the balanced
equation for the reaction between the two
solutions?
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hydrobromic acid
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potassium
hydroxide
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---->
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potassium bromide
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water
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HBr(aq)
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+
KOH(aq)
-
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---->
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KBr(aq)
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+
H2O(l)
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In titration reactions the
mols of H+ must equal the mols of
OH- for a neutral solution. For
monoprotic acids and monohydroxide bases we can use
the equation
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MaVa
= MbVb
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here the
terms are defined as follows
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Ma
is the molarity of the
acid
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Mb
is the molarity of the base
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Va
is the volume of the acid
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Vb
is the volume of the base
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Ma
= 0.20 M HBr ;
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Mb
= 0.11 M KOH;
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Va
= 23.45 mL = 0.02345 L
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Vb
= unknown = X
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MaVa
= MbVb
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(0.20 M HBr )( 0.02345 L ) =
( 0.11 M KOH )( X )
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solve for
volume of the base
Vb
= unknown = X
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X = (0.20 M HBr )( 0.02345 L
)/ (0.11 M KOH) = (0.00469 / 0.11 KOH)
Liter
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X = 0.0426 L ;
X = 0.043
mL rounded off to 2
significant figures because the concentrations have
only two sf.
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|
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