Question

What volume, in milliliters, of 0.120 m naoh should be added to a 0.120 l solution of 0.017 m glycine hydrochloride (p a1=2.350, p a2 = 9.778 ) to adjust the ph to 2.93?

Answer 1

To adjust the pH to 2.93, we can use the Henderson-Hasselbalch equation to calculate the required concentration of glycine hydrochloride and then the volume of 0.120 M NaOH.

Step-by-step explanation:

To calculate the volume of 0.120 M NaOH needed to adjust the pH to 2.93, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

Here, glycine hydrochloride (HA) is a weak acid with two pKa values, and its conjugate base (A-) is glycine. We know the pH, pKa values, and concentration of the glycine hydrochloride solution. We can calculate the required concentration of glycine by rearranging the Henderson-Hasselbalch equation, and then use the dilution equation to find the required volume of 0.120 M NaOH.

Answer 2

To adjust the pH of the solution, 17 mL of 0.120 M NaOH should be added to the solution of 0.120 L of 0.017 M glycine hydrochloride.

Step-by-step explanation:

To adjust the pH of the solution, we need to use the concept of neutralization and calculate the amount of NaOH needed. First, we need to determine the number of moles of glycine hydrochloride in the solution. The molarity of glycine hydrochloride is given as 0.017 M and the volume is given as 0.120 L. Using the formula c = n/V, we can calculate the number of moles of glycine hydrochloride as 0.017 M × 0.120 L = 0.00204 mol.

Next, we need to find the number of moles of NaOH needed to adjust the pH. Since NaOH is a strong base, it will react with the HCl in glycine hydrochloride to form water and salt. From the balanced chemical equation, we know that 1 mol of NaOH reacts with 1 mol of HCl. Therefore, the number of moles of NaOH needed is also 0.00204 mol.

The molarity of NaOH is given as 0.120 M. Using the formula c = n/V, we can calculate the volume of NaOH needed as V = n/c, which is V = 0.00204 mol / 0.120 M = 0.017 L or 17 mL (since 1 L = 1000 mL).