When vanadium(ii) compounds are dissolved in water, the following equilibrium is established. v2+ + > v3+ + 0.5 h2+ oh- what would alter the composition of the equilibrium mixture in favour of the v2+ ions? a. adding an acid b. adding a reagent that selectively precipitates v3+ ions c. allowing the hydrogen to escape as it forms d. making the solution more alkaline

I think the correct answer would be the last option. Making the solution more alkaline would alter the composition of the equilibrium mixture in favor of the vanadium ions. This is based on Le Chatelier's Principle. The statement for the said principle would be "If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change." From the equilibrium reaction, we see that OH- is in the product side, adding more of this ion would shift the equilibrium to the left or forming the reactants which involved the vanadium ions. Adding OH- ions would make the solution more alkaline. 

Your answer is c. explanation: plasma  is the fourth state of matter, derived from the gaseous state of matter.  plasma  can contain free  electrons  and/or  ions. while  plasma is never neutral, it can either be positively or negatively charged.  gases, however, are electrically neutral.