Indicate the concentration of each ion present in the...

Name: indicate the concentration of each ion present in the solution formed by mixing the following assume that the volumes are additive a 30 ml of 0100 m hcl and 100 ml of 0480 m hcl h m cl 45696
Uploaded: 2021-12-24T03:21:04-08:00
Duration: 8 min 17 s
Channel: Ivan Kochetkov
Description: indicate the concentration of each ion present in the solution formed by mixing the following assume that the volumes are additive a 30 ml of 0100 m hcl and 100 ml of 0480 m hcl h m cl 45696

Indicate the concentration of each ion present in the solution formed by mixing the following. (Assume that the volumes are additive.) (a) 30 mL of 0.100 M HCl and 10.0 mL of 0.480 M HCl H+ = ____ M Cl - = ____ M (b) 15.0 mL of 0.350 M Na2SO4 and 27.2 mL of 0.200 M KCl Na+ = ____M K+ = ____M SO42- = ____M Cl - = ____M (c) 3.50 g of NaCl in 46.2 mL of 0.422 M CaCl2 solution Na+ = _____ M Ca2+ = ____M Cl - = _____M

Transcript

00:01 Calculate concentration of which species of which ion after the mixing of the solutions.

00:06 So in the first concentration in the first question, we just mix two different solutions of hcl.

00:13 Therefore, concentration of hydrogen and concentration of chloride are the same because it's just dictated by this technology.

00:22 And here is equals to the total number of most of hcl divided by the total volume.

00:27 So in the denominator we have 30 plus 10 milliliters.

00:34 And then in the nominator we have 30 millimeters by 0 .10 molar plus 10 by 0 .480 molar.

00:48 So here we don't have to convert milliliters per liters because both nominator and denominator have milliliters.

00:57 Now let's calculate this number it equals to 0 .195 molar so that's answer to question a in question b we mix sodium sulfate with potassium chloride so first let's calculate concentration of sodium 2 plus oh sorry sodium plus it equals to 2 because because of the security times 15 .0 millimeters times 0 .350 molar and is divided by the total volume which is 15 .0 plus 27 .2 mm liter.

02:11 Now we can calculate it it equals to 0 .249 molar so that's the concentration of sodium concentration of sulfate equals to the half concentration of sodium which is roughly in this case is roughly 0 125 molar or we can presume that this number is roughly 250 molar so now we have to calculate concentration of potassium and concentration of chlorine concentration of potassium equals to total volume yeah sorry total number of most of potassium chloride divided by total volume or 27 .2 millimeters times 0 .200 molar divided by the total volume.

03:49 Now let's calculate this number.

03:59 It equals to 0129 molar.

04:07 And concentration of potassium equals to the concentration of chloride.

04:13 So therefore concentration of chloride is also the same.

04:17 So that's question be solved...

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