Skip to main content
Ch.16 - Acids and Bases

Chapter 16, Problem 59b

What mass of HI must be present in 0.250 L of solution to obtain a solution with each pH value?

b. pH = 1.75

Verified Solution
Video duration:
3m
This video solution was recommended by our tutors as helpful for the problem above.
108
views
Was this helpful?

Video transcript

Welcome back everyone. How many grams of nitro acid are required to dissolve in 0.350 L of solution to obtain a ph of 1.95. For this problem, we first of all want to recall the concentration or molarity formula. Molarity is equal to the number of moles and divided by the volume in liters V. And moles can be expressed as a ratio of mass lowercase M and molar mass capital M. So if we substitute this expression into the first one, we get molarity equals mass divided by a molar mass divided by volume. And the double division allows us to simplify this to mass divided by the product between molar mass and volume. And since we're looking for grams, we can solve for mass as the product between molarity molar mass and volume. So for this problem, we are given volume, we know the molar mass of nitric acid, we can identify the molar mass using the periodic table. So all that we need is the concentration of nitric acid. Nitric acid is a strong acid. So essentially it ionizes completely. And therefore, based on the stoic geometry of the equation the concentration of hydro would be equal to the concentration of nitric acid itself. And we know that the concentration of hydro is simply expressed as 10 raised to the power of negative ph. And therefore, we can say that the mass would be the concentration of hydro which is sense the power of negative ph multiplied by the molar mass multiplied by volume. And now we know all of these parameters. So we can take sense of the power of negative 1.95 multiplied by the molar mass of nitric acid. That would be 63.01 g per mole. And the volume of this solution is 0.350 L. So if we solve this, we get zero point 2 5 g, that's our final answer. And thank you for watching.