Question #9e3a5

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Doc048
Nov 22, 2017

Working from the following metathesis reaction ...

#2HNO_3(aq) + Ca(OH)_2(aq)# => #Ca(NO_3)_2 + 2H_2O(l)#

The reaction ratio to neutralization is 2 moles #HNO_3# to 1 mole #Ca(OH)_2#

From definition of
#"Molarity"(M) = ("moles"."solute")/("Liters"."solution")#

=> #"moles"."solute"= "Molarity(M)"xx"Volume(L)"#

Therefore, one can use the relationship

1 mole #HNO_3# = 2 moles #Ca(OH)_2#

(Molarity x Volume)#HNO_3# = 2(Molarity x Volume)#Ca(OH)_2#

#(0.177M)HNO_3(aq) xx(28.2ml)HNO_3 "solution")# = #2[Ca(OH)_2(aq)] xx(17.5ml)Ca(OH)_2 "solution")#

Simplifying notations for illustration

(0.177M)(28.2ml) = #2[Ca(OH)_2(aq)]#(17.5ml)

#[Ca(OH)_2]# = #((2(0.177M)(28.2ml))/(17.5ml))#

= #0.570M Ca(OH)_2(aq) "solution"#

Note: In this problem case, the volume values may remain in milliliter dimensions as both sides of the (MV)acid = (MV)base are in ml's.

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