How many grams of KOH are needed to neutralise 500mL of 2mol #L^-1# H2SO4?

1 Answer
Oct 27, 2015

112.22 grams KOH

Explanation:

#color (green) "Step 1:"#

The first thing you need to do is write the chemical equation.

#KOH# + #H_2SO_4# = #K_2SO_4# + #H_2O# (unbalanced)

#color (green) "Step 2:"#

Balance the chemical equation. You need to do this because it's the only way that you can get the "conversion factor" you needed to solve the problem.

#color (red) 2 KOH# + #H_2SO_4# = #K_2SO_4# + #color (red) 2H_2O# (balanced)

left side:

K = 1 x #color (red) 2# = 2
O = (1 x #color (red) 2#) + 4 = 6
H = (1 x #color (red) 2#) + 2 =4
S = 1

right side:

K = 2
O = 4 + (1 x #color (red) 2#) = 2
H = 2 x #color (red) 2# = 4
S = 1

#color (green) "Step 3:"#

Convert the volume of #H_2SO_4# into number of moles. How? By multiplying the given volume to its given molarity. Notice that I already converted the unit from milliliter (mL) to liter (L).

volume #H_2SO_4# = 500 mL = 0.5 L
concentration of #H_2SO_4# = 2M or #2 mol*L^"-1"#

#0.5 cancel L# #H_2SO_4# x #"2 moles"/(1 cancel "L") ##H_2SO_4# = 1 mole #H_2SO_4#

#color (green) "Step 4:"#

Now, going back to the balanced equation,

#color (red) 2 KOH# + #color (red) 1 H_2SO_4# = #color (red) 1 K_2SO_4# + #color (red) 2H_2O#

we know that

#color (red) 1# mole #H_2SO_4# = #color (red) 2# moles #KOH#

Therefore, to convert the computed mole of #H_2SO_4# into mole of #KOH#,

#1# #cancel "mol"##H_2SO_4# x #(2 "mol" KOH)/(1 cancel ("mol") H_2SO_4)# = #2# #"mol"# #KOH#

#color (green) "Step 5:"#

Getting the atomic weights from the periodic table, we can compute the molar mass of #KOH#,

K = #39.10 "g/mol"#
O = #16.00 "g/mol"#
H = #1.01 "g/mol"#

molar mass of #KOH# = #39.10 "g/mol"# + #16.00 "g/mol"# + #1.01 "g/mol"# = #56.11 "g/mol"#

#color (green) "Step 6:"#

Converting the number of moles we got from #color (green) "Step 4"# using the molar mass of #KOH# obtained in #color (green) "Step 5"#,

#2# #cancel "mol"##KOH# x #(56.11 grams KOH)/(1 cancel ("mol") KOH)# = #color (red) (112.22 g KOH)#