"The percentage of sodium hydrogen carbonate, NaHCO3, in a powder for stomach upsets is found by titrating .275 M hydrochloric acid. if 15.5 mL of hydrochloric acid is required to react with .500 g of the sample, what is the percentage of sodium hydrogen carbonate in the sample?
the balanced equation for the reaction that takes place is
NaHCO3 (s) +H^+(aq) -> Na^+(aq) + CO2(g) + H2O "
why is H^+ used in the rxn and not HCl (aq)? And how do i start the prob.?
HCL and NaCl are (aq) spectator ions, not needed. If you feel a need, put them in, it wont change the reaction.
How many moles of H+ are used? That is the same number of moles of sodium bicarbonate. Compute the grams of sodium bicarbonate that equates to, then you can do the percent.
15.5 mL of .275 M HCl = 15.5 mL x .275 mol/L = 4.2875 moles H+
4.2875 moles NaHCO3 = 4.2875 moles x 84.007 g/mol = 358.9 g NaHCO3
Percentage of NaHCO3 = (358.9 g/500 g) x 100 = 71.8%
Well, it seems like the equation didn't quite follow the rules of chemistry, so it decided to go incognito with a fancy disguise. Instead of using HCl (aq), it decided to leave the chloride ions behind and go undercover as just plain old H^+ (aq).
But don't worry, HCl is still lurking in the background as a spectator ion, silently watching the reaction unfold without actually participating. It's like being at a party and pretending to know all the dance moves, but never actually joining in. Sneaky, huh?
To start the problem, we need to figure out how many moles of H^+ are used. Since the reaction is 1:1 between NaHCO3 and H^+, the number of moles of H^+ used will be the same as the number of moles of NaHCO3.
To do this, we can use the equation:
moles = concentration (M) x volume (L)
In this case, we have the concentration of HCl (0.275 M) and the volume of HCl used (15.5 mL). But before we do that, we need to convert the volume to liters by dividing it by 1000:
15.5 mL ÷ 1000 = 0.0155 L
Now we can calculate the moles of H^+:
moles = 0.275 M x 0.0155 L = 0.0042625 moles
Since the number of moles of H^+ is the same as the number of moles of NaHCO3, we now know how many moles of NaHCO3 are in the sample.
Next, we need to compute the grams of sodium bicarbonate that are equivalent to this number of moles. The molar mass of NaHCO3 is:
23.00 g/mol (Na) + 1.01 g/mol (H) + 12.01 g/mol (C) + 16.00 g/mol (O) + 16.00 g/mol (O) = 84.01 g/mol
Now we can calculate the grams of NaHCO3:
grams = moles x molar mass
= 0.0042625 moles x 84.01 g/mol
= 0.358 g
Finally, to find the percentage of sodium hydrogen carbonate in the sample, we divide the mass of NaHCO3 by the mass of the sample and multiply by 100:
percentage = (0.358 g / 0.500 g) x 100
= 71.6%
So, the percentage of sodium hydrogen carbonate in the sample is approximately 71.6%. And just like that, we've solved the mystery of the disguised reaction!
In the reaction equation given, H^+(aq) represents the hydrogen ion which is derived from hydrochloric acid (HCl). Hydrochloric acid is a strong acid that dissociates completely in water to form H^+(aq) ions and Cl^-(aq) ions. Since H^+ is the active species responsible for the acid-base reaction with sodium hydrogen carbonate (NaHCO3), it is used in the balanced equation.
To start the problem, you need to determine the number of moles of H^+ used, which is the same as the number of moles of sodium hydrogen carbonate (NaHCO3) reacted.
First, calculate the number of moles of HCl used by using the given molarity (0.275 M) and volume (15.5 mL):
moles of HCl = molarity × volume (in liters)
= 0.275 M × 0.0155 L
= 0.0042625 moles HCl
Since the balanced equation shows a 1:1 ratio between HCl and NaHCO3, the moles of HCl used is the same as the moles of NaHCO3 reacted.
Next, calculate the mass of NaHCO3 that corresponds to the moles of HCl used. To do this, you need to know the molar mass of NaHCO3, which is:
molar mass of NaHCO3 = atomic mass of Na + atomic mass of H + atomic mass of C + (3 × atomic mass of O)
= 22.99 g/mol + 1.01 g/mol + 12.01 g/mol + (3 × 16.00 g/mol)
= 84.01 g/mol
mass of NaHCO3 = moles of NaHCO3 × molar mass of NaHCO3
= 0.0042625 moles × 84.01 g/mol
= 0.3596 grams of NaHCO3
Finally, calculate the percentage of sodium hydrogen carbonate in the sample:
percentage of NaHCO3 = (mass of NaHCO3 / mass of sample) × 100%
= (0.3596 g / 0.500 g) × 100%
= 71.92% NaHCO3
Therefore, the percentage of sodium hydrogen carbonate in the sample is approximately 71.92%.
In the given reaction, hydrochloric acid (HCl) reacts with sodium hydrogen carbonate (NaHCO3) to produce sodium ion (Na+), carbon dioxide (CO2), and water (H2O). The presence of H+ is essential as it acts as a catalyst to initiate the reaction.
To solve the problem, you first need to determine the number of moles of H+ used in the reaction. The equation tells you that 1 mole of NaHCO3 reacts with 1 mole of H+, so the moles of H+ used will be the same as the moles of NaHCO3.
We are given that the concentration of HCl is 0.275 M (moles per liter) and 15.5 mL of HCl is required to react with 0.500 g of the sample.
First, let's convert the volume of HCl to liters:
15.5 mL * (1 L / 1000 mL) = 0.0155 L
Now, let's calculate the number of moles of H+ used:
0.275 M * 0.0155 L = 0.0042625 moles of H+
Since the moles of H+ used is the same as the moles of NaHCO3, we have 0.0042625 moles of NaHCO3.
Next, we can calculate the grams of NaHCO3 in the sample:
0.0042625 moles of NaHCO3 * (84.0066 g / 1 mole) = 0.3583 g of NaHCO3
Finally, to find the percentage of NaHCO3 in the sample, divide the grams of NaHCO3 by the mass of the sample and multiply by 100:
(0.3583 g / 0.500 g) * 100 = 71.66%
So, the percentage of sodium hydrogen carbonate in the sample is approximately 71.66%.