Asked by MAHLATSE
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
Answers
Answered by
bobpursley
23.73*.01470=25*M
solve for M, the molarity of the gastric juice.
solve for M, the molarity of the gastric juice.
Answered by
drwls
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
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