Fully balanced equation: AgNO3 + NaCl --> AgCl + NaNO3 Net ionic equation: Ag+ +

Question

Fully balanced equation: AgNO3 + NaCl --> AgCl + NaNO3

Net ionic equation: Ag+ +Cl- --> AgCl

Volume of (.0100 M) AgNOs to titrate bay water samples:

March: 7.3 mL

April: 4.46 mL

1) Using the balance equation, calculate the average number of moles of NaCl in each of the titrated samples.

2) Calculate the average numbr of grams of NaCl in each titrated samples.

3) Using the balance equations, calculate the sailinity in ppt of each of the samples. It can be assumed that the density of the bay water is 1 g/mL.

Explanation / Answer

AgNO3+ NaCl--->AgCl+ NaNO3

March condition :

moels of AgNO3 of 0.01M in 7.3 ml = 0.01*7.3/1000=0.000073 moles = moles of NaCl ( from the stoichiometry of the reaction where 1 mole of AgNO3 reacts with 1 mole of NaCl)

In April moles of AgNO3 in 4.46 ml = 0.01*4.46/1000 =0.0000446 moels = moles of NaCl

Average moles of NaCl = (0.000073+ 0.0000446)/2=0.0000588 moles

Mass of NaCl in March =0.000073*(58.5)= 0.004271 gms

Water volume = 7.3 ml = 7.3gml*1 g/ml = 7.3 g

Salinity = 0.004271/7.3 =0.000585 gm/gm =5.85*109ppt

for the second case mass of sdoium chloride =0.0000446 =0.0000446*58.5=0.002609gm85*

0.002609 gmsin 4.46 ml or 4.46 gms of water 0.000585 gm/gm =5.85*109ppt

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