Question #00338

1 Answer
Jul 31, 2017

Answer:

#0.0286M#

Explanation:

We're asked to find the molarity of the #"HCl"# in the gastric juice with some given titration data.

First, let's write the balanced chemical equation for this neutralization reaction:

#"NaOH"(aq) + "HCl"(aq) rarr "NaCl"(aq) + "H"_2"O"(l)#

  • Here's a hint we can use only in the special case of one-to-one molar ratios:

#M_1V_1 = M_2V_2#

where

  • #M_1# and #M_2# are the molarities of the two solutions

  • #V_1# and #V_2# are the volumes of the two solutions

We have:

  • #M_1 = 0.0249M# (#"NaOH"#)

  • #V_1 = 28.7# #"mL"#

  • #M_2 = ?# (#"HCl"#)

  • #V_2 = 25.00# #"mL"#

Plugging in known values, we have

#(0.0249M)(28.7color(white)(l)"mL") = (M_2)(25.00color(white)(l)"mL")#

#M_2 = color(red)(ul(0.0286M#