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Amanda [17]
3 years ago
9

How many milliliters of .45m hcl will neutralize 25ml of 1m koh?

Chemistry
2 answers:
Dennis_Churaev [7]3 years ago
5 0
The  number of  Ml  of 0.45m HCl needed to neutralize  25 ml  of 1 ml KOH  is  calculated  as below

find the moles of KOH  used =  molarity  x volume

= 1  x25 = 25 moles
write the reacting  equation

KOH +HCL = KCl + H2O

by  use   of  mole ratio  between  KOH to HCl  which is 1:1 therefore the  moles of 25 moles

volume of HCl = moles/molarity

 25/0.45 = 55.6 ml  of HCL


solniwko [45]3 years ago
5 0
The balanced equation for the reaction is as follows;
HCl + KOH ---> KCl + H₂O
stoichiometry of HCl to KOH is 1:1
number of KOH moles reacted - 1 mol/L x 0.025 L = 0.025 mol 
according to 1:1 molar ratio 
number of KOH moles reacted = number of HCl moles reacted 
therefore number of HCl moles reacted - 0.025 mol

molarity of HCl is 0.45 M
since 0.45 mol  are in 1000 mL 
therefore 0.025 mol are in a volume of 0.025 mol / 0.45 mol/L  = 55.5 mL 
55.6 mL of HCl is required to neutralise 
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1 year ago
The acid-dissociation constant of hydrocyanic acid (hcn) at 25.0 °c is 4.9 ⋅ 10−10. what is the ph of an aqueous solution of 0.0
vodomira [7]
According to the reaction equation:

and by using ICE table:

              CN-  + H2O ↔ HCN  + OH- 

initial  0.08                        0          0

change -X                        +X          +X

Equ    (0.08-X)                    X            X

so from the equilibrium equation, we can get Ka expression

when Ka = [HCN] [OH-]/[CN-]

when Ka = Kw/Kb

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2 x 10^-5 = X^2 / (0.08 - X)

X= 0.0013

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∴ POH = -㏒[OH]

            = -㏒0.0013

            = 2.886 

∴ PH = 14 - POH

         = 14 - 2.886 = 11.11
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3 years ago
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particles.

just like every dozen contains 12

so

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