a buffer is prepared by combining 10.0g of NaH2PO4 wiht 150mL of 0.20M NaOH and diluting to 2.0L. What is the pH of the buffer, Ka H2PO4-= 6.2x10-8
The reaction between the added solid and NaOH is:
NaH2PO4 + NaOH --> Na2HPO4 + H2O
10.0gNaH2PO4 /120.0g/mol = 0.0833 moles NaH2PO4 initially.
In the rection, the limiting reactant is NaOH.
(0.150 L)(0.20 mol/L) = 0.030 mol NaOH
0.030 moles of NaOH ---> 0.030 moles Na2HPO4 produced, and 0.030 moles NaH2PO4 used up.
After the reaction is complete, 0.0833-0.030 = 0.0533 moles of NaH2PO4 is remaining.
In the final mixture, the acid is H2PO4-(aq). The base is HPO4^2- Use the equation:
pH = pKa + log(Base/Acid)
thanks yall for your help
To find the pH of the buffer, we need to determine the concentration of H2PO4- in the solution.
Step 1: Calculate the moles of NaH2PO4:
Molar mass of NaH2PO4 = 22.99 + 1.01 + (2 * 1.01) + 31.00 + (4 * 16.00) = 119.98 g/mol
Moles of NaH2PO4 = mass / molar mass = 10.0 g / 119.98 g/mol = 0.08334 mol
Step 2: Calculate the initial concentration of H2PO4-:
Initial volume of NaH2PO4 solution = 2.0 L
Concentration of H2PO4- = moles / volume = 0.08334 mol / 2.0 L = 0.04167 M
Step 3: Determine the concentration of H2PO4- after the reaction with NaOH:
NaH2PO4 reacts with NaOH in a 1:1 ratio, so the moles of H2PO4- will be equal to the moles of NaOH used.
Moles of NaOH = concentration * volume = 0.20 M * 0.150 L = 0.030 mol
Concentration of H2PO4- after reaction = (moles - moles of NaOH) / volume = (0.08334 mol - 0.030 mol) / 2.0 L = 0.02667 M
Step 4: Calculate the pKa value for H2PO4-:
pKa = -log(Ka) = -log(6.2 × 10^(-8)) = 7.21
Step 5: Calculate the pH of the buffer:
pH = pKa + log([H2PO4-] / [HPO4^(2-)])
pH = 7.21 + log(0.02667 M / 0.04167 M) = 7.21 + log(0.6397) ≈ 7.21 + (-0.193) = 7.02
Therefore, the pH of the buffer solution is approximately 7.02.
To calculate the pH of the buffer, we need to determine the concentration of the acidic and basic components of the buffer. In this case, the acidic component is NaH2PO4 and the basic component is NaOH.
Step 1: Calculate the concentration of NaH2PO4.
To determine the concentration of NaH2PO4, we first need to convert the mass of NaH2PO4 to moles using its molar mass. NaH2PO4 has a molar mass of 119.98 g/mol.
mass = 10.0 g
molar mass = 119.98 g/mol
moles of NaH2PO4 = mass / molar mass
moles of NaH2PO4 = 10.0 g / 119.98 g/mol
Step 2: Calculate the molarity of NaOH.
We are given that the volume of NaOH used is 150 mL and its concentration is 0.20 M. To calculate the moles of NaOH, we need to convert the volume to liters.
volume = 150 mL
volume in liters = 150 mL / 1000 mL/L
moles of NaOH = molarity x volume in liters
moles of NaOH = 0.20 M x (150 mL / 1000 mL/L)
Step 3: Calculate the total volume and molarity of the buffer solution.
We are given that the buffer solution is diluted to a final volume of 2.0 L. To calculate the molarity of the buffer solution, we need to sum the moles of NaH2PO4 and NaOH and divide by the total volume.
total volume = 2.0 L
total moles = moles of NaH2PO4 + moles of NaOH
molarity of buffer = total moles / total volume
Step 4: Calculate the concentration of HPO4^-2 (the conjugate base).
From the balanced equation of the dissociation of H2PO4^-, we know that one mole of NaH2PO4 produces one mole of HPO4^-2.
concentration of HPO4^-2 = molarity of buffer
Finally, calculate the pH of the buffer using the Henderson-Hasselbalch equation:
pH = pKa + log([HPO4^-2] / [H2PO4^-])
Given that the pKa of H2PO4^- is 6.2 x 10^-8, we can substitute these values into the above equation to calculate the pH of the buffer solution.
Write the equation.
NaH2PO4 + NaOH ==> Na2HPO4 + H2O
Convert 10 g NaH2PO4 to moles.
Convert 150 mL of 0.2 M NaOH to moles.
React and calculate the concn of H2PO4 and HPO4 in the 2.0 L container.
Then pH = pKa + log[(base)/(acid)]
Post your work if you get stuck.