How many mL of a 0.10 M NaOH solution are needed to neutralize 15 mL of a 0.20 M

Question

How many mL of a 0.10 M NaOH solution are needed to neutralize 15 mL of a 0.20 M H_3PO_4 solution? 3NaOH(aq) + H_3PO_4(aq) rightarrow Na_3PO_4(aq) + 3H_2O(l) B. Titration of an Antacid Brand of antacid Bases(s) in antacid Mass of flask Mass of flask and antacid Molarity of HCl solution Total volume (mL) of HCl solution added Molarity of NaOH solution Initial volume (mL) of NaOH solution Final volume (mL) of NaOH solution Mass of antacid Volume of NaOH solution used Volume of excess HCL solution Volume of HCl solution neutralized by antacid mL stomach acid/1 g antacid Write the neutralization equations that take place in the stomach with the base(s) present in the antacid product. How many grams of Mg(OH)_2 will be needed to neutralize 25 mL of stomach acid if stomach acid is 0.10 M HCl?

Explanation / Answer

Q1. No. of moles of NaOH = volume of NaOH (L) x Molarity

= 0.015 x 0.2

= 0.0003 moles

3 NaOH + H3PO4 -> Na3PO4 + 3 H20

1 mole of H3PO4 requires 3 moles of NaOH so,

No. of moles of H3PO4 = 0.0003 x 3 = 0.0009 moles

Volume of NaOH needed = no. of moles/Molarity

= 0.0009/0.1 = 0.009 l

= 9 mL

Q2.  No. of moles of HCl = 0.1 x 0.025 L = 0.0025 moles

2 HCl + Mg(OH)2 -> MgCl2 + 2 H2O

1 mole of Mg(OH)2 requires 2 moles of HCl

No. of moles of Mg(OH)2 = 0.0025/2 = 0.00125 moles

Mass of Mg(OH)2 = 0.00125 x 58.327 = 0.0729 g

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