Verizon LTE 8:00 PM 54% Biochemistry Molecular Weight of NaH2PO4 120 And NaOH mu

Question

Verizon LTE 8:00 PM 54% Biochemistry Molecular Weight of NaH2PO4 120 And NaOH must be added in the ratio of 2:1 base acid to convert 2/3rd of NaH2PO4 to base form. Since the concentration of the buffer is 0.050M, the weight of NaH2PO4 can be found out by the formula- Molarity Weight Molecular weight 0.0S0MWeight 120g/mol Weight- 0.050M x 120g/mol Weight- 6.g To convert 2/3 of the 0.05 M of H2PO4 to HPO4 0.033 M of NaOH is required The stock solution contains 1 mol/L. So 33.3 ml of 1 M NaOH is required. Step 3 0 Was this solution helpful? Chapter 2 Problem 36RE

Explanation / Answer

2/3 of 0.05 = 0.0333

This means that the H2PO4- to convert to HPO4 2- we need 0.0333 M solution of NaOH.

Now M here means molarity. Molarity = n/v

where,

n = number of moles

v = volume in litres. That is moles per liter.

This implies, 0.0333 = n/v

But we have been given a stock solution of 1 Molar. Therefore, we will have to ensure that even if the molarity is different, the number of moles remain the same. Assume v = 1 l.

Therefore 0.0333 = n/v. => n = 0.033 moles. So what volume of 1 Molar stock solution will give us 0.033 moles.

Therefore,

1 = 0.0333/ v

=> v = 0.0333 liter. Therefore, Volume = 33.3 ml of 1M stock solution.

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