7) a solution of HC1/25,00 mL) was titrated to the equivalence point w/ 34.55 mc of...

Question

Question

science chemistry

Answer 1

Answer #1

7)

volume of NaOH = 34.55 mL

molarity of NaOH = 0.1020 M

volume of HCl = 25.00 mL

HCl + NaOH --------------> NaCl + H2O

moles of NaOH = moles of HCl

34.55 x 0.1020 = 25.00 x M

M = 0.1410

concentration of HCl = 0.1410 M

Add a comment

Answer 2

Similar Homework Help Questions

2)a. A 15.00 ml sample of an unknown H2SO4 solution is titrated with a 0.30 M NaOH solution. The equivalence point...

2)a. A 15.00 ml sample of an unknown H2SO4 solution is titrated with a 0.30 M NaOH solution. The equivalence point is reached when 18.00 ml of NaOH solution is added. What is the concentration of the Unknown H2SO4 solution? (1st write balance equation)
A 125 mL solution of sulfuric acid was titrated to the second equivalence point using 25...

A 125 mL solution of sulfuric acid was titrated to the second equivalence point using 25 mL of 0.75 M KOH solution. What is the concentration of the original sulfuric acid solution? Hint: Write the balanced chemical reaction first.
0.450 L of 0.0500 M HCl is titrated to the equivalence point with 39.01 mL of...

0.450 L of 0.0500 M HCl is titrated to the equivalence point with 39.01 mL of a NaOH solution. What is the concentration (in M) of the NaOH solution that was added?

Sulfuric acid (250.0 mL) is titrated with 176.5 mL 2.4 M NaOH to an equivalence point...

Sulfuric acid (250.0 mL) is titrated with 176.5 mL 2.4 M NaOH to an equivalence point (the point where all the sulfuric acid is exactly neutralized). 2 NaOH + H2SO4 → 2 H2O + Na2SO4 (a) How many moles of sulfuric acid were in the original 250.0 mL? (b) What was the concentration of sulfuric acid in the original 250.0 mL sample?
42.04 mL of a solution of the acid HNO3 is titrated, and 52.40 mL of 0.5500-M...

42.04 mL of a solution of the acid HNO3 is titrated, and 52.40 mL of 0.5500-M NaOH is required to reach the equivalence point. Calculate the original concentration of the acid solution.
32.65 mL of a solution of the acid H2C2O4 is titrated, and 43.90 mL of 0.7300-M...

32.65 mL of a solution of the acid H2C2O4 is titrated, and 43.90 mL of 0.7300-M NaOH is required to reach the equivalence point. Calculate the original concentration of the acid solution.

A 25.0 mL NaOH solution of unknown concentration was titrated with a 0.189 M HCl solution....

A 25.0 mL NaOH solution of unknown concentration was titrated with a 0.189 M HCl solution. 19.6 mL HCl was required to reach equivalence point. In a separate titration, a 10.0 mL H3PO4 solution was titrated with the same NaOH solution. This time, 34.9 mL NaOH was required to reach the equivalence point. What is the concentration of the H3PO4 solution?
A 25.0 mL of a solution of acetic acid (HA) of unknown concentration is titrated with...

A 25.0 mL of a solution of acetic acid (HA) of unknown concentration is titrated with 0.1007 M NaOH and the equivalence point volume was determined by graphical means to be 25.42 mL. What is the concentration of the acetic acid? concentration of HA = ___ M
26.24 mL of a solution of the acid H2C2O4 is titrated, and 84.20 mL of 0.4200-M...

26.24 mL of a solution of the acid H2C2O4 is titrated, and 84.20 mL of 0.4200-M NaOH is required to reach the equivalence point. Calculate the original concentration of the acid solution. How many M?

A 25.00 mL sample of hydroiodic acid is titrated to the equivalence point with 38.90 mL...

A 25.00 mL sample of hydroiodic acid is titrated to the equivalence point with 38.90 mL of a 0.125 M barium hydroxide solution. What is the molarity of the acid? (5 points) If the acid had been carbonic acid, instead of hydroiodic acid, but the measurements were the same, what would the carbonic acid’s concentration be? (2 points)