A sample of pure CaCO_3 weighing 0.02500 g is dissolve in hydrochloric acid and

Question

A sample of pure CaCO_3 weighing 0.02500 g is dissolve in hydrochloric acid and the solution diluted to 250.0 ml in volumetric flask. A 25.00 ml adiquet requires ml of an EDTA solution for simulation. Calculate the Titer of the EDTA solution in narrow of ring of CaCO_3/ml of EDTA A 250.0 ml sample of lake water containing Ca^+2 is tilted with 14.00 ml of the above EDTA solution. Calculate the hardness of the water in ring of CaC0_3/ml of lake water IN the spectrophonometric determination of manganese is stored in gram sample of unknown sand dissolved and the manganese considered to MnCl_4. This solution is then diluted to 500 ml in a volumetric flask. The quantity grams of a steel sample, certified to contain 0.25% Mn, is treated in exactly the

Explanation / Answer

1) a) CaCO3 is 0.025g n = m/MW where MW = 60.01g/mol and n = number of mols we have than

n = 0.025g/60.01g/mol = 4.17x10-4mol it are take to 250ml and we know than M = n/V where V is volume in l

M = 4.17x10-4mol/250x10-3l = 1.67x10-3M

C1xV1 = C2xV2 where C2 is the concentration of EDTA and V2 the volume spent of EDTA

C2 = C1xV1/V2 = C2 = 1.67x10-3Mx25ml/43.11ml = 9.68x10-4M

[EDTA] = 9.68x10-4M

b) C1xV1 = C2xV2 =) C2 = C1xV1/V2 = C2 =  9.68x10-4M x 14ml /250ml = 5.42x10-5M of CaCO3

[CaCO3] = 5.42x10-5M in the lake water now make the convertion of M to mg/ml

5.42x10-5M x 60.01g/mol x 1000mg/g x 1l/1000ml = 3.25x10-3mg/ml

2)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.