Final answer:
To neutralize 0.135 L of 0.028 M glycine hydrochloride, you would need 15.12 ml of 0.250 M NaOH, based on the stoichiometry of the neutralization reaction and the molarity of the solutions involved.
Step-by-step explanation:
The student's question focuses on calculating the required volume of a 0.250 M NaOH solution to add to a 0.135 L sample of 0.028 M glycine hydrochloride. To answer this, we need to apply the concept of molarity and the stoichiometry of the reaction between NaOH and glycine hydrochloride, which typically occurs in a 1:1 mole ratio in neutralization reactions.
First, calculate the moles of glycine hydrochloride in the given volume:
0.135 L * 0.028 M = 0.00378 mol.
Since the reaction is 1:1, the same number of moles of NaOH is required to neutralize the glycine hydrochloride.
Using the molarity equation M = moles/L, rearrange to solve for the volume (L) of NaOH solution required:
Volume NaOH = moles NaOH / Molarity NaOH.
Insert the values: 0.00378 mol / 0.250 M = 0.01512 L of NaOH. Convert this to milliliters by multiplying by 1000, resulting in 15.12 ml of 0.250 M NaOH.