23.73*.01470=25*M
solve for M, the molarity of the gastric juice.
iF 23.74Ml OF 0.01470m nAoh are required to completely neutrize 25.00mL of gastric juice,calculate the concentration of the hydrochloric acid(in g/L )in the stomach gastric juice.show steps and calculations
2 answers
You need one mole of NaOH to neutralize one mole of HCl.
You require 0.02374 L x 0.0147 mole/L = 3.49 *10^-4 moles of NaOH
That many moles of HCl have a mass of 1.274*10^-2 g
The concentration is
(1.274*10^-2 g)/(25*10^-3 L)
= 0.510 g/L
You require 0.02374 L x 0.0147 mole/L = 3.49 *10^-4 moles of NaOH
That many moles of HCl have a mass of 1.274*10^-2 g
The concentration is
(1.274*10^-2 g)/(25*10^-3 L)
= 0.510 g/L