(d) What is the molarity of the KOH solution prepared in part (c)...

Question

(d) What is the molarity of the KOH solution prepared in part (c), and how many milliliters of 0.554 M H2SO4 are required to neutralize it?

Accepted Answer

Welcome back everyone. A 9.35 g sample of solid sodium is mixed with 650 g of water at 25 degrees Celsius for the solution, we're going to assume a density of 1.0 g per millimeter at a constant mass. During the reaction, we need to determine the polarity of the sodium hydroxide solution. And what is the volume of 0.5 to five molar hydrochloric acid? And mill leaders required to neutralize the space. We're going to begin by writing out the reaction of solid sodium with water. Recall that sodium is represented as a solid here and we're reacting with water in which we would recognize this as an oxidation reaction reaction where water is oxidizing our sodium and we will form sodium hydroxide as a product as well as hydrogen gas. Now, we need to ensure that this is balanced. So we would place the following coefficients where we have a two in front of our sodium A two in front of our water on the reactive side and on the product side, a two in front of our sodium hydroxide product. Now, with these coefficients, we can confirm that we have a balanced reaction. Now, we're going to need to find our molds of our reactant sodium. So recall that moles is represented by lowercase N. And we would recall that we can find this by taking the molecular or rather the sample mass of our sodium divided by the molecular weight of sodium from the periodic table. And so plugging in what we know our mass of our sodium sample is given in the prompt as 9.35 g divided by the molecular weight of sodium from our periodic table, which we find in group one A which corresponds to the molar mass of 22.99 g per mole. And so we would cancel out our units of grams in the numerator and the denominator here leaving us with moles as our final unit. And we would find That we have 0.4067 moles of sodium that is reacting. Now, we want to figure out how much of our product sodium hydroxide we have produced. So to find our moles of our product sodium hydroxide, We can use our moles of sodium, our reactant, which we just found at 0.4067 moles. And we're going to use the ratio between moles of sodium two moles of our product sodium hydroxide. In which from our balanced equation, we see that we have two moles of sodium equivalent to two moles of sodium hydroxide. So a 2-2 molar ratio. And so when we simplify this, we cancel out moles of sodium. We're left with moles of sodium hydroxide. And we have an amount of our product being 0.4067 moles of sodium hydroxide. Now, we need to next figure out the volume of our solution of water and sodium. But we want to note that from the prompt, we're told that the solution has a constant mass. And so therefore, we would say that the mass of our solution is equal to the mass of water given in the prompt being 650 g. And so to find our volume of our solution, we can take the mass of our solution, which we just noted down divided by the density of our solution in which we're told to assume that it's 1.0 g per meal leader for the density of water. And so we would take 650g divided by our density of 1.00 g per mil leader. We are able to cancel our units of grams leaving us with middle leaders as our final unit of volume. And we would find a volume for the solution being 650 ml. Now, we want this to be in leaders because the molar concentrations of our re agents are understood as moles per liter. And so we're going to multiply by the conversion factor to go from middle leaders in the denominator to leaders in the numerator where we recall that our prefix milli tells us that we have 10 to the negative third of our base unit leaders. And so canceling out our units of milliliters, we results in 0.65 liters as the volume of our solution. So now we're going to need the polarity of our solution in which we can take the moles of sodium hydroxide divided by our volume of our solution, which we just calculated above. And so our moles of sodium hydroxide, we determined above as 0.4067 moles of sodium hydroxide divided by the volume of our solution, which we just determined to be 0.65 liters. Recognizing that we have units of most divided by leaders which we can understand is equal to our molar concentration. We would come up with a molar concentration of 0.6-6 Mohler for our sodium hydroxide solution. So, so far, we've found our first answer and now we want to continue to figure out the volume of our strong acid hydrochloric acid to neutralize sodium hydroxide. So we need to write out a chemical equation for the neutralization of our agents. So the neutralization of sodium hydroxide by hydrochloric acid recall that in a neutralization reaction, our products would be assault being sodium chloride in this case and water. And so now we want to confirm that this is balanced as written and when we analyze our atoms on both sides of the equation, we can confirm that this is a balanced reaction. Now to calculate the moles of hydrochloric acid needed to neutralize our base sodium hydroxide. We would begin with our moles of sodium hydroxide which we determined earlier as .4067 moles of sodium hydroxide which are reacting. And we're going to sorry, not divide, but we're going to multiply by our molar ratio between moles of sodium hydroxide and moles of hydrochloric acid in our balanced equation above. And we can see that we have a 1 to 1 molar ratio based on our coefficients in our equation allowing us to cancel out moles of sodium hydroxide were left with moles of HCL in which we have 0.4067 moles of hydrochloric acid needed to neutralize our sodium hydroxide. And so for our final answer, we need to go ahead and figure out this in terms of volume, so volume of HCL needed. And so we would take our moles of HCL divided by its molar concentration given in the prompt in which we have the moles in the numerator, which we just calculated to be 0.4067 moles of HCL divided by its molar concentration given as in the prompt being 0.5 - five Mueller. And so we would interpret that Mueller concentration as 0.5- moles per liter, which we understand are equal to one moller. And so canceling out moles were left with leaders as our final unit of volume. And we'll find that we have a volume of 0.775 L, which we should convert to mill leaders since the prompt asks us to do so. And so taking 0.775 leaders and actually, we can just keep it here. We're going to multiply by a conversion factor to go from leaders in the denominator to mill leaders in the numerator where we see that for one leader, we have an equivalent of 10 to the third power, male leaders. And so canceling out leaders were left with male leaders as our final unit of volume resulting in a volume of HCL being 775 mL of hydrochloric acid needed to neutralize our solution of sodium hydroxide. So everything highlighted in yellow represents our two final answers to complete this example corresponding to choice C in the multiple choice. I hope everything I explained was clear. If you have any questions, please leave them down below. And I'll see everyone in the next practice video.

Watch next