KPK 11th Class Chemistry Chapter 8 Acids, bases and salts Short Questions Answers
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Class/Subject: 11th Class Chemistry
Chapter Name: Acids, bases and salts
Board: All KPK Boards
- Malakand Board 11th Class Chemistry Chapter 8 Acids, bases and salts short questions Answer
- Mardan Board 11th Class Chemistry Chapter 8 Acids, bases and salts short questions Answer
- Peshawar Board 11th Class Chemistry Chapter 8 Acids, bases and salts short questions Answer
- Swat Board 11th Class Chemistry Chapter 8 Acids, bases and salts short questions Answer
- Dera Ismail Khan Board 11th Class Chemistry Chapter 8 Acids, bases and salts short questions Answer
- Kohat Board 11th Class Chemistry Chapter 8 Acids, bases and salts short questions Answer
- Abbottabad Board 11th Class Chemistry Chapter 8 Acids, bases and salts short questions Answer
- Bannu Board 11th Class Chemistry Chapter 8 Acids, bases and salts short questions Answer
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KPK 11th Class Chemistry Chapter 8 Acids, bases and salts Short Questions Answers
Conjugate acid and base pairs: Let us consider a general reaction of Bronsted acid HA with a base. The reaction can be represented by the following equation. HA + B– A– + HB+ Acid Base conjugate Conjugate Base of HA acid of B In the forward reaction, acid HA donates a proton to base B to form A and HB+ products. In the reverse reaction the HB+ ions is an acid that donates a proton to the A ion, which acts as a base. HCl(g) + N–H3(g) ⇆ N–H4(aq) + Cl–(aq) Acid Base Conjugate Conjugate Acid Base The chloride ion, Cl– is a base by Bronsted definition and is called a conjugate base of hydrochloride acid. According to Bronsted – Lowry “ A reactants and product that differ by a proton[H+] is called conjugate acid – base pair. Every acid has has a conjugate base and every base has a conjugate acid. Thus in an acid base reaction, two conjugate pairs are formed. H2S Weak acid, H2S Weak acid, ionizes in water to produce H+ . Na+ An electron deficient specie, electron pair acceptor ( Lewis acid) SO4-2 electrons pair donor(Lewis base) or proton accepted (Lowry and Bronsted base ) Cl– electron pair doner (Lewis base or Proton acceptor. CO2, H2O, SO2, I–, NH3, OH–, BCl3 SO2 BCl3 I– NH3 OH– As the Ca(OH)2 is a strong base therefore, it ionizes completely in water. So the concentration of OH– is [OH–] = 0.0001 =10-4M We know that pOH = – log[OH–] pOH = – log 10-4 pOH = – (-4) log 10 pOH = + 4 x 1 = 4 We know that pH + pOH =14 pH + 4 = 14 pOH + 4 -4 = 14 – 4 pOH = 10 Buffer action: A buffer solution usually consist of a weakly dissociating acid and the salt of that acid or a weak base and its salt. e.g CH3COOH and CH3COONa, NaHCO3/H2CO3 and NaH2PO4/H3PO4. Suppose a buffer solution of a mixture of acetic acid (CH3COOH)and its salt with a strong base CH3COONa. When small amount of an acid such as HCl is added to this solution , it will combine with the acetate ion to form unionized acetic acid. sodium carbonate(salt) sodium hydroxide carbonic acid The strong base (NaOH) on ionization produces maximum number of OH– as compared to H+ produced by weak acid. Na2CO3 + HOH → NaOH + H2CO3 sodium carbonate(salt) sodium hydroxide carbonic acid The strong base on hydrolysis produces maximum number of OH– as compared to H+ produced by weal acid. b): FeSO4 is a salt salt of a strong acid i.e H2SO4 and weak base i.e Fe(OH)3. The Fe(OH)3 on hydrolysis produces a strong acid i.e H2SO4 which on ionization produces greater number of H+ as compared to the hydroxyl ions produced by weak base. FeSO4 + H2O → H2 SO4 + Fe(OH)3 Ferrous sulphate (salt) strong acid weak base c): NaCl when dissolved in water makes the solution neutral through the following reaction. NaCl → Na+ + Cl– Na+ + HOH → NaOH + H+ Cl– + HOH → HCl + OH– Both NaOH and HCl are strong base and acid, they dissociates to give Na+, OH–, H+ and Cl– ions and the solution behaves neutral
Acids
Bases
BF3 Lewis acid because they are electron deficient specie
NH3 Lewis base because they are electron pair donor
NH4+ Lowry and Bronsted acid (proton donor)
CaO Strong basis because dissolves in water to produces strong base and produces maximum number of OH–
Ag+ An electron deficient specie, an electron pair acceptor (Lewis acid)
KCN on hydrolysis produce a strong base and weak acid. Hence solution contain maximum number of OH–
Lewis acid (electron pair acceptor)
Lewis base ( electron pair donor)
Co2
H2O
Solution:
Solution:
We know that
pH = – log[H+]
-Log [H+] = pH
OR
Log [H+] = – pH
Taking antilog at both side
[H+] = Anti log (-pH)
[H+] = Anti log (-4.87)
[H+] = 0.0001349
OR
[H+] =1.35 x 10-5
We know that
[H+] [OH-] = 10-14
OR
[H+] = 1.35 x 10-5
We know that [H+] [OH-] = 10-14
OR 10-14 1 x 10-14
[OH-] = ___________________ = _______________ = 7.40 x 10-10
. [H+] 1.35 x 10-5
Buffer solutions are capable of maintaining their pH at some fairly constant value even when small amounts of the acid or base are added. Thus a buffer solution is one that tends to maintain its pH when an acid or base is added to it. This property of such a solution is buffer action. The capability of a buffer solution to maintain a definite pH is called buffer capacity.
a): Na2CO3 + HOH → NaOH + H2CO3
