How many milliliters of 0.45M HCI will neutralize 25.0 mL of 1.00M KOH?

1 Answer
Jul 20, 2017

Answer:

#56# #"mL HCl soln"#

Explanation:

We're asked to find the volume (in #"mL"#) of a #0.45M# #"HCl soln"# that will neutralize #25.0# #"mL"# of a #1.00M# #"KOH soln"#.

To do this, let's first write the chemical equation for the reaction:

#"HCl"(aq) + "KOH"(aq) rarr "KCl"(aq) + "H"_2"O"(l)#

What we can do is use the molarity equation to find the number of moles of #"KOH"# that are present:

#"molarity" = "mol solute"/"L soln"#

#"mol solute" = ("molarity")("L soln")#

#"mol KOH" = (1.00"mol"/(cancel("L")))overbrace((0.0250cancel("L")))^"converted to liters" = color(red)(0.0250# #color(red)("mol KOH"#

Now, we can use the coefficients of the chemical equation to find the relative number of moles of #"HCl"#:

#color(red)(0.0250)cancel(color(red)("mol KOH"))((1color(white)(l)"mol HCl")/(1cancel("mol KOH"))) = color(red)(0.0250# #color(red)("mol HCl"#

Lastly, we can use the molarity equation again to find the volume of #"HCl"# solution:

#"molarity" = "mol solute"/"L soln"#

#"L soln" = "mol solute"/"molarity"#

#"L HCl soln" = (color(red)(0.0250)cancel(color(red)("mol HCl")))/(0.45(cancel("mol"))/"L") = color(blue)(0.056# #color(blue)("L"# #= color(blue)(ul(56color(white)(l)"mL"#