The formula for the monoprotic acid is taken as HA, reaction with base is as follows;
HA + NaOH ---> NaA + H₂O
Stoichiometry of acid to base is 1:1
At the neutralisation point, number of HA moles = number of base moles
Number of NaOH moles reacted = 0.100M / 1000 mL /L x 30.0 mL = 0.003 mol
Therefore number of HA moles reacted = 0.003 mol
the mass of acid 0.384 g
Therefore molar mass - 0.384 g/ 0.003 mol = 128 g/mol
0.0760 m
do this by:
finding the moles of NaOH which will be <span>5.702 E -3 m
</span>
next find the moles of H3PO4 which will be <span>1.90 E -3 m</span><span>
calulcate </span>25 ml sample molarity = 0.07603 m, just put 0.0760<span>
</span>
.............................
<span>HNO2 =====> H+ + NO2-
</span>I<span>nitial concentration</span> = 0.311
<span>C = -x,x,x </span>
<span>E = 0.311-x,x,x
</span>KNO2 ====>K+ + NO2-
<span>Initial concentration = 0.189 </span>
<span>C= -0.189,0.189,0.189 </span>
E = 0,0.189,0.189