Sodium hydroxide is used extensively in acid-base titrations because it is a strong inexpensive base. A sodium hydroxide solution was standardized by titrating 27 mL of 0.1628 M standard hydrochloric acid. The initial buret reading of the sodium hydroxide was 2.44 mL, and the final reading was 39.39 mL. What was the molarity of the base solution?
HCl + NaOH ==> NaCl + H2O
Since this is a 1:1 reaction, one may use this formula.
mL x M = mL x M
mL NaOH = 39.39 - 2.44
What is an isotopes
To determine the molarity of the sodium hydroxide solution, we can use the equation:
M1V1 = M2V2
Where:
M1 = Molarity of the hydrochloric acid
V1 = Volume of the hydrochloric acid (in liters)
M2 = Molarity of the sodium hydroxide solution
V2 = Volume of the sodium hydroxide solution (in liters)
Let's start by converting the volumes from milliliters to liters:
V1 (hydrochloric acid) = 27 mL = 27/1000 L = 0.027 L
V2 (sodium hydroxide) = (39.39 mL - 2.44 mL) = 36.95 mL = 36.95/1000 L = 0.03695 L
Now, we can plug in the values into the equation:
M1 * V1 = M2 * V2
Substituting the values we have:
0.1628 M * 0.027 L = M2 * 0.03695 L
Now, solve for M2:
M2 = (0.1628 M * 0.027 L) / 0.03695 L
M2 = 0.1199 M
Therefore, the molarity of the sodium hydroxide solution is approximately 0.1199 M.