1. How many mL of 0.100 M NaOH are required to neutralize 25.0 mL of 0.15 M H3PO
ID: 551429 • Letter: 1
Question
1. How many mL of 0.100 M NaOH are required to neutralize 25.0 mL of 0.15 M H3PO4? Write th balance acid base neutalization reaction to start.
2. Toxic chromate can be precipitated from an aqueous solution by bubbling SO2(g) through the solution. How many grams of SO2 are required to treat 1.0 x 10^8 L of 0.000100 M CR+3?
3. Predict the products of the reaction and write its balanced molecular equation, ionic equation and net ioniv equation.
BaCl2 + H3PO4 ----->
4. Identify the reducing agent and the oxidizing agent in the equation. And balance the redox reaction in acidic medium.
CrO4(s) + I^- (aq) ---> Cr^+3(aq) + I2(aq)
oxidizing agent:
Reducing agent:
Oxidation half reaction:
Reduction half reaction:
Explanation / Answer
1. Use the relation M1V1 = M2V2
Here, M1 and M2 are initial and final molarities; V1, V2 are the initial and final volumes.
Substitute and calculate the initial volume as follows:
V1 = M2V2/V1
= 0.15 M* 25.0 mL/0.100 M
= 37.5 mL
Therefore, the required volume is 37.5 mL
The balanced equation is: H3PO4(aq) + 3 NaOH(aq) = Na3PO4(aq) + 3 H2O(l)
As per chegg guidelines, first question is answered.
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