The following equation is given:

S2O3 -2 + 2H +1 ---> S + SO2 + H2O

[HCl] = 2 mol/L
[Na2S2O3]= 0.3 mol/L

what is the concentration (mol/L) of H+ if you were to add:

a) 20 mL of HCL with 5 mL Na2S2O3 & with 5 mL H2O

b) 15 mL of HCl with 5 mL Na2S2O3 & 10 mL H2O

c) 7.50 mL of HCl with 5 mL Na2S2O3 & 17.50 mL H2O

d) 5 mL of HCl with 5 mL Na2S2O3 & 20 mL H2O

Why did the chemist open a bakery? Because he kneaded a change of pace!

a) To solve for the concentration of H+, we need to determine the moles of HCl and H2O added, as well as the initial concentration of H+. Since you added 20 mL of HCl, you can convert this volume to liters by dividing by 1000 (since there are 1000 mL in a liter), giving you 0.020 L. The moles of HCl can be calculated by multiplying the volume in liters by the concentration, which gives you: 2 mol/L * 0.020 L = 0.040 mol. Similarly, for H2O, you added 5 mL, which is equivalent to 0.005 L. However, since H2O is a solvent and not involved in the reaction, its concentration doesn't change, so we'll assume it's still 1 mol/L. Therefore, the concentration of H+ is 2 mol/L.

b) This equation is really "pouring" in the variables! For this scenario, you added 15 mL of HCl, which is 0.015 L. Multiplying this by the concentration of HCl, we get: 2 mol/L * 0.015 L = 0.030 mol. You also added 10 mL of H2O, which is 0.010 L. Remember, the concentration of H2O doesn't change, so it's still 1 mol/L. Therefore, the concentration of H+ is still 2 mol/L.

c) Are you "diluting" with anticipation? In this case, you added 7.50 mL of HCl, which is 0.0075 L. Multiplying by the concentration, we get: 2 mol/L * 0.0075 L = 0.015 mol. You also added 17.50 mL of H2O, which is 0.0175 L. And just like before, the concentration of H2O remains unchanged at 1 mol/L. Thus, the concentration of H+ is still 2 mol/L.

d) This equation is turning up the "amphiphilic" comedy! Is it "aqueous" to say that? You added 5 mL of HCl, which is 0.005 L. So, multiplying by the concentration of HCl, we have: 2 mol/L * 0.005 L = 0.010 mol. And since you added 20 mL of H2O, which is 0.020 L, the concentration of H2O still remains 1 mol/L. Therefore, the concentration of H+ is still 2 mol/L.

So no matter how you mix it, the concentration of H+ remains a constant 2 mol/L. It seems like someone forgot to shake things up in this reaction!

To find the concentration of H+ in each scenario, we need to use the concept of stoichiometry and the given concentrations of [HCl] and [Na2S2O3]. Here's how to calculate the concentration of H+ for each scenario:

a) 20 mL of HCl with 5 mL Na2S2O3 and 5 mL H2O:

Step 1: Convert the volumes to L:
20 mL = 20/1000 = 0.02 L
5 mL = 5/1000 = 0.005 L

Step 2: Calculate the moles of HCl:
moles of HCl = volume of HCl (L) × [HCl]
moles of HCl = 0.02 L × 2 mol/L = 0.04 mol

Step 3: Calculate the moles of Na2S2O3:
moles of Na2S2O3 = volume of Na2S2O3 (L) × [Na2S2O3]
moles of Na2S2O3 = 0.005 L × 0.3 mol/L = 0.0015 mol

Step 4: According to the balanced equation, the ratio of HCl to H+ is 2:2. Therefore, the moles of H+ produced are the same as the moles of HCl.

Step 5: Calculate the total volume of the solution:
total volume = volume of HCl + volume of Na2S2O3 + volume of H2O
total volume = 0.02 L + 0.005 L + 0.005 L = 0.03 L

Step 6: Calculate the concentration of H+:
[H+] = moles of H+ / total volume
[H+] = 0.04 mol / 0.03 L = 1.33 mol/L

b) 15 mL of HCl with 5 mL Na2S2O3 and 10 mL H2O:
Follow the same steps as in scenario (a) and calculate the concentration of H+.

c) 7.50 mL of HCl with 5 mL Na2S2O3 and 17.50 mL H2O:
Follow the same steps as in scenario (a) and calculate the concentration of H+.

d) 5 mL of HCl with 5 mL Na2S2O3 and 20 mL H2O:
Follow the same steps as in scenario (a) and calculate the concentration of H+.

By performing the calculations, you will obtain the concentrations of H+ for each scenario.