AE signal for solution containing Na+ was 4.27 mV. If 5.00 mL of 2.08 M NaCl...
AE signal for solution containing Na+ was 4.27 mV. If 5.00 mL of 2.08 M NaCl is added to 95.0 mL of the initial solution the signal is measured as 7.98 mV. What is the conc of Na in the original sample?
Homework Answers
Signal 1= 4.27 mV
Signal 2 = 7.98 mV
volume of sample = 5.0 ml
volume of dilution = 95 ml
total volume = 100 ml
concentration of = 2.08M
[Na+] / 0.104 + 0.950 [Na+] = 4.27/7.98
where 0.104 = concentration x volume of sample/ total volume = 2.08 x 5/100 and 0.950 = volume of dilution / total volume = 95/100
[Na+] 7.98 = 4.27 ( 0.104 + 0.950 [Na+])
[Na+] 7.98 = 0.44408 + 4.0565 [Na+]
[Na+] 7.98 -4.0565 [Na+] =0.44408
3.9235 [Na+] = 0.44408
[Na+] initial concentration = 0.44408/3.9235 = 0.11318 M
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