Excess Ag+ is added to 50 mL of NaBr soln and precipitated out 22.8 mg of AgBr. What is the original concentration of Br- ion in the solution in ppm?

Convert 22.8 mg AgBr to moles. moles = grams/molar mass.

Since 1 Br^- is in 1 molecule of AgBr, that must be the moles Br^- in the 50 mL.

Convert moles Br^- to grams in the 50 mL.
Now I would convert that to mg/L which will make it in ppm.

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To find the original concentration of Br- ion in the solution in parts per million (ppm), we need to know the moles of AgBr precipitated and the volume of the solution.

First, let's calculate the moles of AgBr precipitated. We know the mass of AgBr is 22.8 mg. The molar mass of AgBr is the sum of the atomic masses of silver (Ag) and bromine (Br), which is approximately 187 g/mol.

Moles of AgBr = mass of AgBr / molar mass of AgBr
= 22.8 mg / 187 g/mol
≈ 0.122 moles

Next, we need to find the volume of the solution. It is given that the solution is 50 mL, so the volume is 50 mL = 0.05 L.

Now, we can calculate the concentration of Br- ion in the solution in moles per liter (M).

Concentration (M) = Moles / Volume
= 0.122 moles / 0.05 L
= 2.44 M

Finally, to convert this concentration to parts per million (ppm), we multiply by the conversion factor 1,000,000.

Concentration (ppm) = Concentration (M) x 1,000,000
= 2.44 M x 1,000,000
= 2,440,000 ppm

Therefore, the original concentration of Br- ion in the solution is 2,440,000 ppm.