1.) It takes 2.085x10^1 mL of 1.30x10^-1 M NaOH to titrate 2.9660x10^-1 grams of solid monoprotic acid (HX). What is the molar mass of this acid. Express in scientific not. w/ 3 sig figs.
Here is my work, but I'm not sure if it's right. Please check. Thanks.
(1.30x10^-1 M NaOH * 2.085x10^1 mL )/1000
=0.0027105
2.9660x10^-1 grams/0.0027105= molar mass= 109.426=1.09x10^2
2.) 2.72x10^-1 grams of solid monoprotic acid (HX) w/ mm of 1.0572x10^2 g/mol was titrated w/ 1.220x10^-1 M NaOH. How many mL of NaOH would it take to reach the endpoint?
Work:
39.997 mm of NaOH
2.72x10^-1 grams / 39.997g NaOH= 0.00680051 moles/1.0572x10^2 g/mol * 1000 ml = 6.43x10^-8
are these correct?
So I double checked, the first is right. Although I'm not sure about 2.)
1 is right.
2 is wrong. In 2 you divided grams HX by molar mass NaOH to find mols
I say 1 is right. I just checked the procedure and not the math.
Let's break down each question and go through the steps to see if the calculations are correct.
1.) Molar mass of HX:
To calculate the molar mass of HX, we can use the formula:
Molar mass = mass of HX (in grams) / moles of HX
Given:
- Volume of NaOH used (V) = 2.085x10^1 mL
- Concentration of NaOH (C) = 1.30x10^-1 M
- Mass of HX = 2.9660x10^-1 grams
First, convert the volume of NaOH from milliliters (mL) to liters (L):
V (in L) = 2.085x10^1 mL / 1000 = 2.085x10^-2 L
To determine the moles of NaOH, use the equation:
moles of NaOH = concentration of NaOH (in mol/L) * volume of NaOH (in L)
moles of NaOH = 1.30x10^-1 M * 2.085x10^-2 L = 2.7003x10^-3 mol NaOH
Since NaOH and HX react in a 1:1 ratio, the moles of HX will be equal to the moles of NaOH:
moles of HX = 2.7003x10^-3 mol HX
Now, calculate the molar mass of HX:
molar mass of HX = mass of HX (in grams) / moles of HX
molar mass of HX = 2.9660x10^-1 grams / 2.7003x10^-3 mol ≈ 109.861 g/mol
Rounding to three significant figures, the molar mass of HX is 1.10x10^2 g/mol.
2.) Volume of NaOH:
To determine the volume of NaOH needed to reach the endpoint, we can use the equation:
moles of NaOH = mass of HX (in grams) / molar mass of HX
Given:
- Mass of HX = 2.72x10^-1 grams
- Molar mass of HX = 1.0572x10^2 g/mol
moles of HX = 2.72x10^-1 grams / 1.0572x10^2 g/mol
moles of HX ≈ 2.573x10^-3 mol
Since NaOH and HX react in a 1:1 ratio, the moles of NaOH will also be 2.573x10^-3 mol.
Now, to determine the volume of NaOH:
volume of NaOH = moles of NaOH / concentration of NaOH (in mol/L)
Given:
- Concentration of NaOH = 1.220x10^-1 M
volume of NaOH = 2.573x10^-3 mol / 1.220x10^-1 M
volume of NaOH ≈ 2.108x10^-2 L ≈ 21.08 mL
So, the volume of NaOH required to reach the endpoint is approximately 21.08 mL.
Based on the calculations, the molar mass for the first question is 1.10x10^2 g/mol, and the volume of NaOH for the second question is 21.08 mL.