MOLARITY Problem!?

Question

Use this equation: HCl (aq) + NaHCO3(aw) = H2O(l) + CO2(g) +NaCl(aq)

A) If 1.87g NaHCO3 is added to 350mL of 0.100M HCl, calculate the molarity of H+ AFTER the rxn goes to completion.

B) What volume of .950M KOH would be required to neutralize the remaining HCl in (a)?

Sorry! 1 tablet contain 1.87g NaHCO3.

Answer 1

A) 1.87 g NaHCO3 = 1.87g * (mole/84g)=
0.0222 mole

350ml*0.1moleH+/1000ml= 0.035 mole H+

H+ left = 0.035 - 0.0222 = 0.0128 moles

the molarity is in 350ml soln = 0.0128/.35 = 0.036M

B) xml *0.950moleOH/1000ml = 0.0128 mole H+

solving x = 13.47 ml

you'll need 13.47 ml of 0.95M NaOH soln to neutralize whats left.

Answer 2

Lancenigo di Villorba (TV), Italy

Your question in a FUNNY one!

ABOUT HYDROCHLORIC ACID & ITS IONS
Hydrochloric Acid is a gas when it is manipulated in Pure Form at Room Temperature. It results very soluble in water forming strong acidic solutions.
(WARNING!! Its solutions may be corrosive versus skin, eyes, mouth! Don't ingest them! Its solutions may lift up some fumes, don't inhalate them! Don't add water to concentrated solutions! Don't expose these solutions to heat or external energy's sources).
Since it is a STRONG ACID, you may assume hydrochloric undergo COMPLETE DISSOCIATION in its formed ions, e.g. hydrogen and chloride's one.

ABOUT CARBONIC ACID & ITS IONS
Carbonic Acid is not a stable stuff in room conditions because its trend involves the decomposition toward Carbon Dioxide bubbles. Physico-chemical data about the CO2's dissolution permit you to know the water-solubilities of this gas and you can assume that the dissolute form become Carbonic Acid. The latter behaves as a WEAK ACID and a BI-BASIC ACID meaning that it take part to two Acid-Base's Reactions
Since it is a WEAK ACID, anyone of the following reactions obeys to an "Acid-Base Reaction's Constant", e.g. in aqueous media the well-known "Acidity's Constant" stated at 25°C

4.5E-7 = Ka1 = |HCO3-| * |H+| / |H2CO3|
5.6E-7 = Ka2 = |CO3--| * |H+| / |HCO3-|

QUESTION a)
Starting from the stiochiometric relationship, ONE MOLE OF NaHCO3 REACTS DESTROYING ONE MOLE OF HCl OR "H+ ION".
(NaHCO3's moles) = 1.87 / 84 = 2.2E.2 mol
(HCl's mole) = 350E-3 * 0.100 = 3.5E-2 mol
(HCl's residuals) = 3.5E-2 - 2.2E-2 = 1.3E-2 mol
(|HCl| Final Molarity) = 1.3E-2 / 350E-3 = 3.7E-2 M

QUESTION b)
You execute an "Acid-Base's Determination" moving toward the "Equivalence Point" where Acid's amount is determined as similar to the Base's amount dosed.
(KOH's amount) = (HCL's residuals) = 1.3E-2 mol
(KOH's dose) = 1.3E-2 / 0.95 = 1.37E-2 L = 13.7 ml

I hope this helps you.