NaHCO3 + HCl -> NaCl+H2O+CO2
What is the normality of a sodium bi carbonate solution containing 5.08g NaHCO3 in 150 ml solution?
Normality? H+ eqivalents is 2, so
Normality=molesNaHCO3/.150 * 2
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
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.
.403?
Yes, 0.403 N.
To find the normality (N) of a solution, you need to know the number of equivalents (eq) of the solute and the volume of the solution.
In this case, the equation is NaHCO3 + HCl -> NaCl + H2O + CO2. From the equation, we can see that 1 mole of NaHCO3 reacts with 1 mole of HCl, and 1 mole of NaHCO3 gives 1 equivalent of NaHCO3.
First, calculate the number of moles of NaHCO3 in the solution using its molar mass. NaHCO3 has a molar mass of 84 g/mol.
Number of moles of NaHCO3 = Mass of NaHCO3 / Molar mass of NaHCO3 = 5.08 g / 84 g/mol = 0.06 mol
Since 1 mole of NaHCO3 gives 1 equivalent of NaHCO3, the number of equivalents is also 0.06 eq.
Next, convert the volume of the solution from milliliters (ml) to liters (L). There are 1000 ml in 1 L.
Volume of solution = 150 ml / 1000 = 0.15 L
Finally, calculate the normality using the formula:
N = Number of equivalents / Volume of solution
N = 0.06 eq / 0.15 L = 0.4 N
Therefore, the normality of the sodium bicarbonate solution containing 5.08 g NaHCO3 in 150 ml is 0.4 N.