Ostwald Dilution Law Mathematical Interpretation
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Aqueous solution of HCl has the pH = 4. Its molarity would be
0.0001 M
10 M
0.4 M
4 M
The dissociation constant of two acids HA1 and HA2 are 3.14 × 10−4 and 1.96 × 10−5 respectively. The relative strength of the acids will be approximately
1:4
4:1
1:16
16:1
Nicotinic acid (Ka = 1.4 x 10−5) is represented by the formula HNic. For a solution which contains 0.10 mol of nicotinic acid per 2.0 litre of solution, the percent dissociation is:
4.2%
12.77%
1.66%
3.33%
- 1.4×10−5
- 1.4×10−4
- 3.7×10−4
- 2.8×10−4
What is the conc. of acetic acid that must be added to 0.5M HCOOH solution so that the dissociation of both acids is the same? [An example of Isohydric solutions] Ka of acetic acid 1.8×10−5, Ka of formic acid = 2.4×10−4
4.5 M
3.2 M
6.66 M
12.3 M
- 1.4×10−5
- 1.4×10−4
- 3.7×10−4
- 2.8×10−4
- 9.6×10−3
- 2.1×10−4
- 1.25×10−6
- 4.8×10−5
What is the conc. of acetic acid that must be added to 0.5M HCOOH solution so that the dissociation of both acids is the same? [An example of Isohydric solutions] Ka of acetic acid 1.8×10−5, Ka of formic acid = 2.4×10−4
4.5 M
3.2 M
6.66 M
12.3 M
Ka=1.8×10−5 for CH3COOH
Take (1.34)2=1.8
- 3 mM HCOOH(Ka=6×10−4)
- 0.1 M CH3COONa
- 1.34 mM HCl
- 0.1 M CH3COOH
Ka=1.8×10−5 for CH3COOH
Take (1.34)2=1.8
- 3 mM HCOOH(Ka=6×10−4)
- 0.1 M CH3COONa
- 1.34 mM HCl
- 0.1 M CH3COOH
- 2×10−6 mole L−1
- 2×10−5 mole L−1
- 3×10−6 mole L−1
- 3×10−5 mole L−1