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.
For the reaction \( A + B \to C \), the rate law is found to be \( \text{rate} = k[A]^2[B] \). If the concentration of \( A \) is doubled and \( B \) is halved, by what factor does the rate change?
In the given circuit, if the potential at point B is 24 V, the potential at point A is:
