A 500.0-mL buffer solution is 0.100 M in HNO₂ and 0.150 M in KNO₂. Determine if each addition would exceed the capacity of the buffer to neutralize it. a. 250 mg NaOH

Video transcript

Hi everyone. So I asked if the addition of 1.5 g of potassium hydroxide will exceed the capacity of the 250 millimeter buffer solution. That is 0.125 moles of hydrogen fluoride. 0.175 moles of sodium fluoride. We know that potassium hydroxide is it base I was gonna consume the acid and it will exceed it. The mall's potassium hydroxide is greater than the malls of hydrogen fluoride. So we need to first calculate the moles of E at 1.5 g of potassium hydroxide. And in one moment we have the smaller mouse. This is 39.098 g. That's 15 .999 g. That's 1.008 g. Let's give us 56 11 g. So you're going to get 0.0267 moles of potassium hydroxide. We know that the polarity equal some balls of the salute lot by leaders. And the solution we need to convert our volume into leaders. 250 male leaders. And in one leader we have 1000 ml. We're gonna get 0.25 leaders. 0.125 malls. Oh how's your floor ride in one leader? And then we have 0.25 L of solution. We're gonna get 0.03 balls of hydrogen fluoride. And this is more than a number of multiple tasks and hydroxide and now it will not exceed the capacity. Thanks for watching my video And I hope it was helpful