Acidic acid solutions called cleaning vinegar are effective in dissolving calcium carbonate deposits on shower tiles what sample size of 0.918 molar cleaning vinegar it would require 27.50 mL of 0.5185 molars of NAOH for neutralization
To determine the amount of cleaning vinegar required for neutralization, we can use the concept of stoichiometry. Here, the balanced chemical equation is:
CH3COOH + NaOH → CH3COONa + H2O
From the balanced equation, we can see that 1 mole of acetic acid (CH3COOH) reacts with 1 mole of sodium hydroxide (NaOH) to form 1 mole of water (H2O).
First, we need to find out how many moles of NaOH are in 27.50 mL of 0.5185 M solution:
moles of NaOH = molarity × volume (in L)
moles of NaOH = 0.5185 mol/L × (27.50 mL / 1000 mL/L) = 0.014259 mol
Since the molar ratio is 1:1, we need an equal amount of moles of acetic acid:
moles of acetic acid = 0.014259 mol (1 mol CH3COOH / 1 mol NaOH) = 0.014259 mol
Now we can find the volume of 0.918 M cleaning vinegar that contains this amount of acetic acid:
volume (in L) = moles of acetic acid / molarity of cleaning vinegar
volume (in L) = 0.014259 mol / 0.918 mol/L = 0.015537 L
Now convert this volume to milliliters:
volume (in mL) = 0.015537 L × 1000 mL/L = 15.54 mL (approximately)
Therefore, it would require 15.54 mL of 0.918 M cleaning vinegar for neutralization.