I need to calculate the amount of NaH2PO4 (in grams) and 1.00M NaOH solution (in ml) needed to make 250 ml of 0.20M H2PO4- at pH 8.00.
My work is lengthy and involved but the answer I got was 7.00 ml of the NaOH and 0.0070 moles of the NaH2PO4 needed.
Can you tell me if I am correct? If not I will retry the calculation.
See your post above.
To determine if your calculations are correct, let's break it down step by step.
First, let's look at the balanced chemical equation for the reaction between NaH2PO4 and NaOH:
NaH2PO4 + NaOH -> Na2HPO4 + H2O
We can see that one mole of NaH2PO4 reacts with one mole of NaOH to form one mole of Na2HPO4.
Now, let's calculate the moles of H2PO4- needed to make 250 ml of 0.20M solution. The molarity (M) is expressed in moles per liter (mol/L), so for 0.20M H2PO4-, we have:
0.20 mol/L * 0.250 L = 0.05 moles of H2PO4-
Since the reaction between NaH2PO4 and NaOH is 1:1, we need 0.05 moles of NaH2PO4.
To convert this to grams, we need to know the molar mass of NaH2PO4. The atomic masses are:
Na: 22.99 g/mol
H: 1.01 g/mol
P: 30.97 g/mol
O: 16.00 g/mol
Adding these up:
1 * Na + 2 * H + 1 * P + 4 * O = 22.99 + 2.02 + 30.97 + 64.00 = 119.98 g/mol
To calculate the mass of NaH2PO4 needed, we use:
Mass = Moles * Molar mass
Mass = 0.05 moles * 119.98 g/mol = 5.999 g
So, you are correct that approximately 6 grams of NaH2PO4 is needed.
Now let's calculate the volume of 1.00M NaOH solution needed to reach pH 8.00. The pH value is a measure of the concentration of hydrogen ions (H+). Since pH 8.00 is slightly basic, it means it has a lower concentration of H+. To determine the volume of NaOH, we need to know the balanced equation and the reaction that occurs between H2PO4- (from NaH2PO4) and NaOH:
H2PO4- + OH- -> HPO42- + H2O
Since it is a monoprotic acid, we can assume that 1 mole of H2PO4- reacts with 1 mole of OH- to form 1 mole of HPO42-.
To reach pH 8.00, we need to neutralize H+ ions equivalent to the concentration of OH- ions in NaOH.
The concentration of OH- ions in the NaOH solution is:
1.00 mol/L
To neutralize the H+ ions, we can equate the concentration of OH- ions to the concentration of H+ ions caused by the dissociation of H2PO4-:
[OH-] = [H+]
Since the pH is 8.00, we can calculate the concentration of H+ ions using the equation:
[H+] = 10^(-pH) = 10^(-8.00) = 1.0 x 10^(-8) mol/L
We now have the concentration of H+ ions. Since the reaction between H2PO4- and OH- is 1:1, the concentration of H+ ions is equal to the concentration of OH- ions in NaOH.
Concentration of OH- ions = 1.0 x 10^(-8) mol/L
Now, to determine the volume of NaOH (in mL) needed to reach the desired pH, we'll use the equation:
Volume (mL) = Moles / (Concentration × 1000)
Volume of NaOH = 0.05 moles / (1.0 x 10^(-8) mol/L × 1000) = 5.0 mL
Therefore, your calculation for the volume of NaOH needed is incorrect. The correct volume is 5.0 mL, not 7.00 mL.
In summary, you need approximately 6 grams of NaH2PO4 and 5.0 mL of 1.00M NaOH solution to make 250 mL of 0.20M H2PO4- at pH 8.00.