What volume of 0.125M H2SO4 is required to neutralize 2.5g of Ca (OH)2?

2 Answers
Mar 10, 2018

Well, we address the equation...

#Ca(OH)_2(s) +H_2SO_4(aq) rarr CaSO_4(s)darr + 2H_2O(l)#

Explanation:

#"Moles of calcium hydroxide"=(2.5*g)/(74.1*g*mol^-1)=0.0337*mol#

And thus we need ONE equiv of sulfuric acid...

#=(0.0337*mol)/(0.125*mol*L^-1)xx1000*mL*L^-1=270*mL#

Mar 10, 2018

#270.27# #mL#

Explanation:

Write out the complete and balanced equation of the reaction in order to find the ratio of moles required

#H_2SO_4(aq) + Ca(OH)_2(aq) → CaSO_4(s)+2H_2O(l)#

State symbols in the equation above

#s# - Solid
#l# - Liquid
#aq# - Aqueous (dissolved in water)

Relative Formula Mass of Calcium Hydroxide #(Ca(OH)_2)#

#Ca# = #40#
#O# = #16#
#H#= #1#

#40#x#1# = #40#
#16#x#2# = #32#
#1#x#2# = #2#

#40+32+2 = 74#
#RFM# = #74gmol^-1#

To find the number of moles of Calcium Hydroxide #(Ca(OH)_2)# when we have the mass and relative formula mass we use the formula and substitute in the values

#n = m/M#

Where

#n# = number of moles (#mol#)
#m# = mass (#grams#)
#M# = relative formula mass (#gmol^-1#)

#n# = #2.5 / 74#

Number of moles in (#1#) Calcium Hydroxide = #5/148#

Since the reactants are in a #1:1# Ratio we can state that Sulfuric Acid #(H_2SO_4)# and Calcium Hydroxide #(Ca(OH)_2)# have the same number of moles.

Number of moles in (#1#) Calcium Hydroxide #(Ca(OH)_2)# and (#1#) Sulfuric Acid

#∴# = #5/148#

Now that we have the number of moles and the concentration of Sulfuric Acid #(H_2SO_4)# we can find the volume.

To do this use the formula

#n# = #c# x #v/1000#

Where

#n# = number of moles (#mol#)
#c# = concentration (#moldm^-3#)
#v# = volume (#cm^3#)

Substitute in the values and rearrange for # v#

#5/148##mol# = #0.125##moldm^-3*v/1000#

#5/148mol#x#1000# = #0.125##moldm^-3*v#

#1250/37mol# = #0.125##moldm^-3\8v#

#(1250/37mol)/( 0.125moldm^-3) = v#

#v# = #270.27# #cm^3#

To mL

#v# = #270.27##mL#

#:.# the volume of #0.125M# #H_2SO_4# required to neutralize #2.5g# of #Ca(OH)_2# is #270.27mL#