Normality? H+ eqivalents is 2, so
Normality=molesNaHCO3/.150 * 2
NaHCO3 + HCl -> NaCl+H2O+CO2
What is the normality of a sodium bi carbonate solution containing 5.08g NaHCO3 in 150 ml solution?
8 answers
How do i find the moles of NaHCO3?
do you add all the atomic masses.....giving you 84.004?
1120.05
or is it ......
5.08NaHCO3/150ml*1000ml/1L*2eq/84.004g NaHCO3 = 0.806 N
5.08NaHCO3/150ml*1000ml/1L*2eq/84.004g NaHCO3 = 0.806 N
The definition of normality is number of milliequivalents/mL (or number equivalents/L).
Equivalent weight for NaHCO3 in the reaction shown = 23+1+12+48 = about 84 but you need to verify that and do it exactly. I'm just rounding the atomic weights to the nearest whole number.
So 5.08/84 = number of equivalents.
Then number of equivalents/0.150 L = normality = 0.403 N. Your answer of 0.806 is high by a factor of 2.
Equivalent weight for NaHCO3 in the reaction shown = 23+1+12+48 = about 84 but you need to verify that and do it exactly. I'm just rounding the atomic weights to the nearest whole number.
So 5.08/84 = number of equivalents.
Then number of equivalents/0.150 L = normality = 0.403 N. Your answer of 0.806 is high by a factor of 2.
.403?
Yes, 0.403 N.
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