Use Le' Chatelier's Principle.

this is a solution at equilibrium:
2CrO4^2-(aq)+ 2H^+(aq)< >Cr2O7^2-(aq) +H20

2CrO4^2- yellow
Cr2O7^2- orange

I just have to make predictions of the colour changes when:

a)Add 0.3 M NaOH drop to 5 drops of 0.3 M K2CrO4

b)Add 0.3 M HCl by drop to 5 drops 0.3 M K2CrO4

c)Add 0.3 M N HCl drop to 5 drops of 0.3 M K2CrO7

d)Add 0.3 M NaOH drop to 5 drops of 0.3 M K2CrO7

I think it's:
a)yellow
b)yellow
c)orange
d)orange

I'm really confused please help!What is the difference when you add NaOH and HCl?

Look at the equilibrium reaction. It involves H^+.

So for example, we take K2CrO4 and add H^+ so it drives the reaction to the right, changes CrO4^-2 (which is yellow) to Cr2O4^-2 (which is orange).

Or if we take Cr2O7^-2 (orange) and add NaOH. The NaOH reacts with the H^+, which is on the left side of the equation, that reduces H^+, that shifts the equilibrium to the left and makes CrO4^-2 so the solution turns from orange to yellow.

so lets say for c. HCl adds H+ ions to the product side which makes it more acidic and therefore yellow?

Adding HCl adds H^+ and that drives the reaction to the right. What product and what color are formed on the right?

so i am a bit confused it mainly depends on whether you add H+ (from HCl) or OH- (from NaOH)?

it does not matter whether u add it to CrO4^-2 or Cr2O7^-2?

Because when you add HCl it reacts with H+ which is on the left so the equilibrium moves towards the left and hence yellow

similarly when u add NaOH it reacts with H+ to produce H20. Whne you add it to Cr2O7^-2 the solution turns yellow but when u add it to CrO4^-2 the solution turns orange?

I think you're on the right track because the amount of H+ or OH- affects the equilibrium, which relates to the principle

your wronggg

To predict the color changes in each scenario, we can apply Le Chatelier's Principle. This principle states that when a system at equilibrium is subjected to a stress, it will shift in a way that minimizes the effect of that stress.

Now let's analyze each scenario:

a) Adding 0.3 M NaOH drop to 5 drops of 0.3 M K2CrO4:
In this case, NaOH is a base and it will increase the concentration of hydroxide ions (OH-) in the solution. According to Le Chatelier's Principle, the system will shift to the left to counteract the increase in OH-. This means the concentration of CrO4^2- ions will increase, resulting in a more intense yellow color. Therefore, the solution will become more yellow.

b) Adding 0.3 M HCl by drop to 5 drops of 0.3 M K2CrO4:
HCl is an acid, and when it is added, it will increase the concentration of H+ ions in the solution. According to Le Chatelier's Principle, the system will shift to the right to counteract the increase in H+. This means the concentration of Cr2O7^2- ions will increase, resulting in a more intense orange color. Therefore, the solution will become more orange.

c) Adding 0.3 M NH4Cl drop to 5 drops of 0.3 M K2CrO7:
NH4Cl is also an acid, so it will increase the concentration of H+ ions in the solution. Following Le Chatelier's Principle, the system will shift to the left to counteract the increase in H+. This means the concentration of Cr2O7^2- ions will decrease, resulting in a less intense orange color. Therefore, the solution will become less orange.

d) Adding 0.3 M NaOH drop to 5 drops of 0.3 M K2CrO7:
NaOH is a base, and it will increase the concentration of OH- ions in the solution. According to Le Chatelier's Principle, the system will shift to the right to counteract the increase in OH-. This means the concentration of Cr2O7^2- ions will increase, resulting in a more intense orange color. Therefore, the solution will become more orange.

In summary:
a) The solution will become more yellow.
b) The solution will become more orange.
c) The solution will become less orange.
d) The solution will become more orange.

The difference between adding NaOH and HCl lies in their respective effects on the concentration of H+ and OH- ions in the solution. NaOH increases the concentration of OH- ions, while HCl increases the concentration of H+ ions. These changes in ion concentrations result in different shifts in the equilibrium position of the reaction, leading to different color changes.