Calculating Phpreap

More reactions of Acids & Bases

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Neutralization Reaction = acid + base → salt + water – The salt comes from the cation of the base & the anion of the acid. – Water is made from the H+ on the acid & the OHon the base.

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Molecules with OH- can be acidic or amphoteric depending on the other oxygens present. More oxygens take electrons away from the O-H bond and make it easier to break (more polar/more acidic)

Water  Water

is polar--it has a positive and a negative end. – Water can form OH- and H+ – H+ reacts with water to form H3O+ (hydronium ion) – This can be written as the self ionization of water: 2H2O → H3O+ + OH-

More on Water  

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In pure water only 1 x 10-7 M of H+ and OHform. Pure water is neutral. You can multiply the concentration of H+/H3O+ and OH- to get the ion product constant for water Kw. Kw= 1 x 10–14 M2 [H+] [OH-1] = Kw [H+ or H3O +] [OH-1] = 1 x 10–14 M2 From this equation, it is possible to figure out either [H+1] or [OH-1] if the other one is known.

Sample Problem 

If the [OH-] in a solution is 1.15 x 10-12 M, what is the [H3O+] of this solution?

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[H3O+][OH-]= 1 x 10-14 M2

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[H3O+] = Kw [OH-] [H3O+] = 1 x 10-14 M2 = 0.00870 M 1.15 x 10-12 M

Your Turn!  If

the [H3O+] in a solution is 2.98 x 10-5 M, what is the [OH-] of this solution?  [OH-] = Kw [H3O+] [OH-] = 1 x 10-14 M2 = 3.36 x 10-10 M 2.98 x 10-5 M

Definitions    

An acidic solution is one in which the [H3O+] is greater than the [OH-]. An basic solution is one in which the [OH-] is greater than the [H3O+]. pH was developed to be an easier way to calculate acid & base concentration than Kw. The pH of a solution is the negative logarithm of the hydrogen ion concentration. pH = -log[H+] or pH = -log[H3O+]

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The pOH of a solution is the negative logarithm of the hydroxide ion concentration. pOH = -log[OH-]

pH and pOH  In

a neutral solution, [H3O+] = 1 x 10-7 M. The pH of the solution is 7.0  The pH of pure water or a neutral solution is 7.0  A neutral solution has a pOH of 7.0  Simple Relationship: pH + pOH = 14

Sample Problem  What

is the pH of a HBr solution with a hydrogen-ion concentration of 1.36 x 10-4 M? pOH? pH = -log[H3O+] = -log(1.36 x 10-4 ) = -(-3.87) = 3.87 pH + pOH = 14 pOH = 14 – 3.87 = 10.13

Your Problem  What

is the pH of a HCl solution with a hydrogen-ion concentration of 9.85 x 10-2 M? pOH? pH = -log[H3O+] = -log(9.85 x 10-2) = 1.01 pOH = 14 – 1.01 = 12.99

Try another one!  What

is the pOH of a NaOH solution with a hydroxide-ion concentration of 6.35 x 103 M? pH? Is this an acidic, basic or neutral solution? pOH = -log[OH-] = -log(6.35 x 10-3) = -(-2.20) = 2.20 pH = 14 – 2.20 = 11.80 pH = 11.80 = Basic

Another one  What

is the pH of a KOH solution with a hydroxide-ion concentration of 3.45 x 10-4 M? pOH = -log[OH-] = -log(3.45 x 10-4 ) = 3.46 pH = 14 - pOH = 14 – 3.46 pH = 10.54

Almost Done! 

What is the [OH-] and [H+] concentration of an HI solution with a pH of 4.5? pH = -log[H3O+] log[H3O+] = -pH pH = 4.5 log[H3O+] = -4.5 [H3O+] = antilog - 4.5 or 10-4.5

[H3O+] =3.16 x 10-5 M [OH-] = 1 x 10-14 M2 = 3.16 x 10-10 M 3.16 x 10-5 M

Last one! 

What is the [OH-] and [H3O+] concentration of a LiOH solution with a pOH of 3.6? pOH = -log[OH-] log[OH-] = -pOH pOH = 3.6 log[OH-] = -3.6 [OH-] = antilog – 3.6 or 10-3.6

[OH-] = 2.51 x 10-4 M [H3O+] = 1 x 10-14 M2 = 3.98 x 10-11 M 2.51 x 10-4 M

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