5. The glass electrode used in commercial pH meters gives an electrical response proportional to the concentration of hydrogen ion. To convert these responses to a pH reading, the electrode must be calibrated against standard solutions of known H+ concentration. Determine the weight in grams of sodium dihydrogen phosphate (NaH2PO4ƒ| H2O; FW 138) and disodium hydrogen phosphate (Na2HPO4ƒ| H2O; FW 142) needed to prepare 1L of a standard buffer at pH 7.0 with a total phosphate concentration of 0.1 M. pKa of phosphoric acid 6.86.
Did you hear about the chemist who tried to make a buffer solution? He ended up getting stuck in a pH jam!
But don't worry, I'm here to help! Let's calculate the weight of sodium dihydrogen phosphate (NaH2PO4•H2O) and disodium hydrogen phosphate (Na2HPO4•H2O) needed for your standard buffer.
First, we need to determine the ratio of the two components in the buffer. The Henderson-Hasselbalch equation can help us with that:
pH = pKa + log([A-]/[HA])
Since the buffer pH is 7.0 and the pKa is 6.86, we can rearrange the equation to find the ratio:
[A-]/[HA] = 10^(pH - pKa)
[A-]/[HA] = 10^(7.0 - 6.86)
[A-]/[HA] ≈ 1.23
Now, let's consider the total phosphate concentration. Since the concentration of NaH2PO4 and Na2HPO4 will be the same in the buffer, each compound will contribute half to the total phosphate concentration.
0.1 M total phosphate concentration ÷ 2 ≈ 0.05 M for each compound.
The molecular weights of NaH2PO4•H2O and Na2HPO4•H2O are 138 g/mol and 142 g/mol, respectively.
To calculate the weight of each compound needed, we can use the formula:
Weight = (concentration in moles) × (molecular weight)
Weight of NaH2PO4•H2O = (0.05 mol/L) × (138 g/mol) × (1 L) = 6.9 g
Weight of Na2HPO4•H2O = (0.05 mol/L) × (142 g/mol) × (1 L) = 7.1 g
So, approximately 6.9 grams of sodium dihydrogen phosphate and 7.1 grams of disodium hydrogen phosphate are needed to prepare 1 liter of the standard buffer with a pH of 7.0 and a total phosphate concentration of 0.1 M.
I hope these calculations don't make your pH levels rise!
To determine the weight of sodium dihydrogen phosphate (NaH2PO4·H2O) and disodium hydrogen phosphate (Na2HPO4·H2O) needed to prepare a 1L standard buffer at pH 7.0 with a total phosphate concentration of 0.1 M, we can use the Henderson-Hasselbalch equation and the known pKa of phosphoric acid.
First, let's understand the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where:
- pH is the desired pH value (in this case, 7.0)
- pKa is the negative logarithm of the acid dissociation constant (given as 6.86 for phosphoric acid)
- [A-] is the concentration of the conjugate base (disodium hydrogen phosphate, Na2HPO4)
- [HA] is the concentration of the acid (sodium dihydrogen phosphate, NaH2PO4)
Since we want to prepare a buffer solution with a pH of 7.0, we can substitute the values into the Henderson-Hasselbalch equation:
7.0 = 6.86 + log([Na2HPO4]/[NaH2PO4])
To simplify the calculation, we need to know the ratio of [Na2HPO4] to [NaH2PO4]. We can denote this ratio as r:
[Na2HPO4]/[NaH2PO4] = r
Now, we can rewrite the Henderson-Hasselbalch equation using the concentration of total phosphate (0.1 M) and the ratio (r):
7.0 = 6.86 + log(r * [NaH2PO4] / [NaH2PO4])
Simplifying further:
0.14 = log(r)
Now, let's find the value of r by using the known pKa (6.86) and rearranging the equation:
r = 10^(pH - pKa)
r = 10^(7.0 - 6.86)
r = 10^0.14
r ≈ 1.38
Now that we have the value of r, we can determine the concentration of NaH2PO4 and Na2HPO4 in the buffer solution.
Let's assume x is the concentration of NaH2PO4. Therefore, the concentration of Na2HPO4 will be 1.38 times x.
The total concentration of phosphate is 0.1 M. Thus, the sum of the concentration of NaH2PO4 and Na2HPO4 should be equal to that.
x + 1.38x = 0.1
2.38x = 0.1
x ≈ 0.042 M
Now, we can calculate the moles of each compound needed to prepare 1L of the buffer solution:
Moles of NaH2PO4 = concentration (M) * volume (L)
Moles of NaH2PO4 = 0.042 * 1
Moles of NaH2PO4 = 0.042 moles
Moles of Na2HPO4 = (1.38 * 0.042)
Moles of Na2HPO4 = 0.058 moles
Finally, we can determine the weight of each compound needed using their respective molecular weights (FW):
Weight of NaH2PO4 = moles * FW
Weight of NaH2PO4 = 0.042 * 138
Weight of NaH2PO4 ≈ 5.796 grams
Weight of Na2HPO4 = moles * FW
Weight of Na2HPO4 = 0.058 * 142
Weight of Na2HPO4 ≈ 8.236 grams
Therefore, to prepare 1L of the standard buffer at pH 7.0 with a total phosphate concentration of 0.1 M, you would need approximately 5.796 grams of sodium dihydrogen phosphate (NaH2PO4·H2O) and 8.236 grams of disodium hydrogen phosphate (Na2HPO4·H2O).