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The pH of glycine at the first half equivalence point is 2.34 and that at second half equivalence point is 9.60.At the equivalence point (The first inflection point) The pH is :
pH = pKa1 = 2.34.
pH = pKa2 = 9.6.
pH = = 5.97.
A 1.458 g of Mg reacts with 80.0 ml of a HCI solution whose pH is –0.477. The change in pH after all Mg has reacted. (Assume constant volume. Mg = 24.3 g/mol.)(log 3 = 0.477)
Mg(aq) + 2HCl(aq) → MgCl2 (aq) + H2
Millimoles of HCl = 3 × 80 = 240 mM
Moles of HCl after reaction = 240 – 60 × 2 = 120
New Molarity = = 1.5 M
pH = – log[H+] = – log 1.5 = – 0.176
Change is pH = – 0.176 – (– 0.477) = 0.3
Find the pH (initial pH –final pH) when 100 ml 0.01 M HCl is added in a solution containig 0.1 m molesof NaHCO3 solution of negligible volume ( Kai =10–7, Ka, =10–11 for H2CO3) :
pH of NaHCO3 solution = 9
Now H+ + HCO3– → H2CO3
∴ no. of mmole of HCl remaining = 1 – 0.1 = 0.9 mmole
∴ pH = – log (9 × 10–3) = – 2 log 3 + 3
The ionization constant of benzoic acid is 6.46 x 10–5 and Kc for silver benzoate is 2.5 x 10–13. How many times silver benzoate is more soluble in a buffer of pH = 3.19 as compared to its solubility in pure water ?
In pure water: C6H5COOAg C6H5COO– + Ag+
(s – x) s
C6H5COO– + H2O C6H5COOH + OH–
(s – x) x x
s (s – x) = 2.5 × 10–13 ..……….
.………. (2)
Calculate s.
In buffer: C6H5COOAg C6H5COO– + Ag+
(s’ – x’) s’
C6H5COO– + H+ C6H5COOH
S’ (s’ – x’) = 2.5 × 10–13.………. (3)
= 6.46 × 10–5 .………. (4)
solve for s’.
Calculate
30 ml of 0.06 M solution of the protonated form of an anion acid methonine (H2A+) is treated with 0.09 MNaOH. Calculate pH after addition of 20 ml of base. pKa 1, = 2.28 and pKa2 = 9.2.
pH = = = 5.74
A certain acid–base indicator is red in acid solution and blue in basic solution 75% of the indicator is presentin the solution in its blue form at pH = 5. Calculate the pH at which the indicator shows 90% red form?
pH = pK1 + log
5 = pK1 + log
⇒ pK1 = 4.523
⇒ K1 = 3 × 10–5
pH = 4.523 + log = 4.523 – 0.954 = 3.56
Calculate the molar solubility of zinc tetrathiocyanato–N–mercurate (II) if its Ksp = 2.2 x 10–7.
Zn [Hg(NCS)4] Zn+2 + [Hg(NCS)4]2–
⇒ KSP = S2
⇒ S = = 4.69 × 10–4 mol/L
An acid–base indicator which is a weak acid has a pKa value = 5.45. At what cocentration ratio of sodiumacetate to acetic acid would the indicator show a colour half–way between those of its acid and conjugate base forms? pKa of acetic acid = 4.75. [log 2 = 0.3]
Indicator is weak acid HIN H+ + In–
given it shows colour at half way of ionization
Ka = so pKa = pKa = pH = 5.45
but for CH3COOh and CH3 COONa buffer
pH = pKa + log
5.45 = 4.75 + log
⇒ =
The indicator constant of phenolphthalein is approximately 10–10. A solution is prepared by adding 100.01c.c. of 0.01 N sodium hydroxide to 100.00 c.c. of 0.01N hydrochloric acid. If a few drops of phenolphthalein are now added, what fraction of the indicator is converted to its coloured form?
NaOH + HCl → NaCl + H2O
mm 100.01 × 0.01 100 × 0.01
conc. 0 1 1
HPh H+ + Ph–
Ka =
⇒ 10–10 = =
so .
A certain mixture of HCl and CH3 – COOH is 0.1 M in each of the acids. 20 ml of this solution is titrated against 0.1M NaOH. By how many units does the pH change from the start to the stage when the HCl is almost completely neutralised? Ka for acetic acid = 1.8 x 10–6.
Initial [H+] = 0.1 (H+ from weak acid can neglect)
after neutralization of HCl concentration of CH3COOH = = 0.05
volume would double[H+] =
[H+] = = 9.48 × 10–4
pH = 3.03
change in pH unit = 3.03 – 1 = 2.03
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