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A. Match each type of titration to its pH at the equivalence point. Weak acid, strong base Strong acid, strong base...

A. Match each type of titration to its pH at the equivalence point.
Weak acid, strong base
Strong acid, strong base
Weak base, strong acid

pH less than 7
pH equal to 7
pH greater than 7

B. A 56.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 28.0 mL of KOH.

C. Consider the titration of 50.0 mL of 0.20 M NH3 (Kb=1.8 x 10^-5) with 0.20 M HNO3. Calculate the pH after addition of 50.0 mL of the titrant.

D. A 30.0 mL volume of 0.50 M CH3COOH (Ka = 1.8 x 10^-5) was titrated with 0.50 M NaOH. Calculate the pH after addition of 30.0 mL of NaOH.
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Answer #1
Concepts and reason

Solutions are classified into acids or bases based on the pH value.

Titration is a process where a known concentrated solution is used for determining the concentration of an unknown solution. Solution of a known concentration is called as titrant while solution of an unknown concentrated is called as an analyte.

pH is a measure of hydrogen ion present in a solution.It is the negative logarithm of the H+ concentration to the base 10. It gives the acidic strength to a solution.

Fundamentals

Acid: Acid substance is the one which can donate a proton or accept a pair of electron.

Base: Base substance is the one which can accept a proton or donate a pair of electrons.

Solutions are classified into acids or bases based on the pH value. If pH value is less than 7, the solution is acidic, and if pH is greater than 7, the solution is basic.

pH of a solution can be determined by identifying the concentration of in the solution.

The pH value is negative logarithm (base10 ) of the Hydrogen ion concentration[н] .

pH log,

Molarity: Molarity is the ratio of the number of moles to the volume of the solution.

Number of moles Molarity (M)Volume of solution(V)(Units: moles/:Liter) Number of moles Molarity (M)x Volume of solution (V) O

KH0OH 1x1014 Here [H0lis the hydronium ion concentration [OH Jis the hydroxide ion concentration

Equilibrium constant is expressed asK . К, is the expression for the base dissociation or ionization constant for an equilibrium reaction. К, for equilibrium reaction is expressed as follows:

ОН + B* ВОН Гон Тв] К [Вн]

K. is the expression for an acid dissociation or ionization constant for an equilibrium reaction. K. for equilibrium reaction is expressed as follows:

NH4 NH H NH,H NH, К.

(A)

Weak acid, strong base = pH greater than 7 Strong acid, strong base pH equal to7 Weak acid, strong base = pH less than 7

(B)

Given: Volume of HBr= 56.0mL 0.056L Concentration of HBr = 0.25M

Concentration of KOH = 0.50M Volume of KOH=28.0mL - 0.028L

The reaction between KOH and HCl is as follows КОН (а) + нсІ(аq)—>н,о(аq) + кCI

Concentration of OH = Concentration of KOH Moles of OH Concentration of KOHxVolume of KOH =0.50M 0.028L 0.014mol Concentratio

Moles of H Concentration of HClxVolume of HCl - 0.25Mx0.056L 0.014mol Moles of OH ion is equal to moles of H* ion. Therefore,

(C)

Given: Volume of NH 50.0mL =0.050L Concentration of NH, = 0.20M 18x10 Ky of NH

Concentration of HNO, 0.20M Volume of HNO = 50.0mL =0.050L

The reaction is: NH, (aq) H (aq)NH, moles of NH, Concentration of NH,xVolume of NH, = 0.050L*0.2M -0.01mol

Concentration of H* x Volume of H moles of H 0.050L*0.2M 0.01mol moles of NH, and moles of H are equal. Hence, it is the equ

Total volume of solution = 0.05L 0.05L 0.1L

moles of NH Concentration of NH Total volume of solution 0.01mol 0.1L = 0.1M

К,К, - К. w к, к, К, 1x101 1.8х105 %3D 5.6х1010

NH H* NH 1(м) С (м) Е(м) 0.1 0 0 X X 0.1-х х X

NH, NH [NH,H NH, H* к, 5.6x1010 (x)(x) (0.1-х) +

Neglect x in the denominator and assume x is very small as K, is very smal 5.6x10 10 0.1 x2 5.6x10 X 7.48x10

[H] pH= -log (*) pH log 7.48x10 = 5.13

Ans: Part A

Weak acid, strong base = pH greater than 7 Strong acid, strong base pH equal to7 Weak acid, strong base = pH less than 7

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