When sodium chloride is added to a saturated solution of lead(II) chloride, some

Question

When sodium chloride is added to a saturated solution of lead(II) chloride, some of the lead(II) chloride precipitates. That phenomenon is called__the common ion effect. selective recitation. a solubility anomaly. super saturation. deionization. In a titration of strong monoprotic acids and bases, there is large change in pH_when the concentration of acid is exactly equal to the concentration of base. at the point where pH - pKa of the acid. when the volume of acid is exactly equal to the volume of base. when the number of moles of acid is exactly equal to the number of moles of base. at the point where pH - pKb of the base. Buffer is: A neutralized salt Solution of a weak acid or base and its conjugate partner acid The equilibrium of a weak acid The equilibrium of a weak base The equilibrium of a strong Iodine-131 has a half-life of 8.1 days and is used as a tracer for the thyroid gland. If a patient drinks a sodium iodide (NaI) solution containing iodine- 131 on a Tuesday, how many days will it take for the concentration of iodine- 131 to drop to 10.0% of its initial concentration? What is the concentration of [H^+] in a 0.10 M solution of acetic acid? The K_a value for acetic acid is 1.8 times 10^-5 What is the solubility of barium sulfate in a solution containing 0.010 M sodium sulfate? The k_sp value for barium sulfate is 1.1 times 10^-10 pH = - log[H^+] pOH = - log[OH^-] K_w = [H_3O^+][OH^-] = 1 times 10^-14 K_sp = [A+][B^-] In N_t/N_0 = - 0.693 t/t_1/2

Explanation / Answer

18) we are adding one common ion (iodide ion) which will

19) at point of equivalence there will be sudden change in pH

The equivalence point is reached when the moles of acid = moles of base (mono acid and mono base)

20) Buffer is mixture of weak acid + its salt with strong base

Or weak base + its salt with strong acid

Examp;e: CH3COOH + CH3COONa

NH4OH + NH4Cl

1) we know that half life = t1/2 = 0.693 / K

K = rate constant = 0.693 / t1/2 = 0.693 / 8.1 = 0.0856 day^-1

Now we will use first order rate law

time = 2.303 / K ] log [initial concentration / final concentration]

time = 2.303 / 0.0856 [ log (A0 / 0.1 A0]

time = 26.9 days

2) the formula is

Ka = [H+]^2 / [acid]

Ka = 1.8 X 10^-5

1.8 X 10^-5 X 0.1 = [H+]^2

[H+] = 1.34 X 10^-3

pH = -log[H+] = 2.872

3) Ksp = [ Ba+2][SO4-2]

Na2SO4 wil give 0.01 M of SO4-2

Ksp = 1.1 X 10^-10 = [Ba+2] [ SO4-2 + 0.01]

[Ba+2] = [SO4-2] = solubility of BaSO4

Which is negligible as compared to 0.01

1.1 X 10^-10 = [Ba+2] [ 0.01]

[Ba+2] = 1.1 X 10^-8 = solubility of Ba+2

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