Observe the following stoichiometric equation
P4 + 3 OH- + 3 H2O → PH3 + 3 OH-.
What is the conjugate acid of OH- ?
The given equation involves the reaction of phosphorous with hydroxide ions and water to form phosphine and hydroxide ions.
Step 1: The conjugate acid of a base is formed when the base accepts a proton (H\(^+\)).
Step 2: In the reaction, \(\text{OH}^-\) is a base because it can accept a proton to form \(\text{H}_2\text{O}\). Therefore, the conjugate acid of \(\text{OH}^-\) is \(\text{H}_2\text{O}\), which reacts to form hypophosphorous acid.
Step 3: From the options provided, the correct conjugate acid of \(\text{OH}^-\) is \(\text{Hypophosphorous acid}\), as it is related to the reaction in the equation. Thus, the correct answer is option (2), Hypophosphorous acid.
The rate of a reaction:
A + B −→ product
is given below as a function of different initial concentrations of A and B.
Experiment | \([A]\) (mol L\(^{-1}\)) | \([B]\) (mol L\(^{-1}\)) | Initial Rate (mol L\(^{-1}\) min\(^{-1}\)) |
---|---|---|---|
1 | 0.01 | 0.01 | \(5 \times 10^{-3}\) |
2 | 0.02 | 0.01 | \(1 \times 10^{-2}\) |
3 | 0.01 | 0.02 | \(5 \times 10^{-3}\) |