Determine the volume (in mL) of 0.844 M hydrochloric acid (HCl) that must be added to 38.0 mL of 0.650 M sodium hydrogen phosphate (Na2HPO4) to yield a pH of 6.92.
Assume the 5% approximation is valid and report your answer to 3 significant figures. Look up pKa values.
I would use the Henderson-Hasselbalch equation, pH = pKa + log[(base)/(acid)]\
HPO4= + H+ ==> H2PO4-
38 mL HPO4= x 0.650M = 24.7 mmoles.
To determine the volume of hydrochloric acid (HCl) that needs to be added, we need to follow these steps:
1. Write the balanced chemical equation for the reaction between HCl and Na2HPO4:
HCl + Na2HPO4 → NaH2PO4 + NaCl
2. Identify the significant compounds and ions present in the solution. In this case, we have HCl, Na2HPO4, NaH2PO4, and NaCl.
3. Determine the initial concentrations of each compound using their molarity (M) and volumes given:
Initial concentration of HCl: 0.844 M (unknown volume)
Initial concentration of Na2HPO4: 0.650 M (38.0 mL)
4. Calculate the initial moles of Na2HPO4 by multiplying the initial concentration by the volume:
Moles of Na2HPO4 = Initial concentration × Volume
= 0.650 M × 0.0380 L (since volume is given in mL)
= 0.02470 moles
5. Since Na2HPO4 is a weak base, it will react with HCl to form NaH2PO4. Find the amount of HCl needed to react with Na2HPO4 by using the stoichiometry of the balanced equation. The 1:1 molar ratio between HCl and Na2HPO4 indicates that the same number of moles will react.
Moles of HCl needed = Moles of Na2HPO4
= 0.02470 moles
6. Now we need to convert the moles of HCl needed to volume. Use the equation:
Moles of HCl needed = Concentration of HCl × Volume of HCl
Volume of HCl = Moles of HCl needed / Concentration of HCl
= 0.02470 moles / 0.844 M
= 0.02924 L (conversion from moles to L)
7. Finally, convert the volume to milliliters (mL) by multiplying by 1000:
Volume of HCl in mL = 0.02924 L × 1000 mL/L
= 29.24 mL
Hence, the volume of 0.844 M HCl that needs to be added to 38.0 mL of 0.650 M Na2HPO4 to yield a pH of 6.92 is approximately 29.24 mL.