A current of 11.3 A is applied to 1.25 L of a solution of 0.552 M...

Question

A current of 11.3 A is applied to 1.25 L of a solution of 0.552 M HBr converting some of the H+ to H2(g), which bubles out of solution. What is the pH of the solution after 73 minutes?

Accepted Answer

Hey everyone in this example, we're told that when a current is passed through a hydrochloric acid solution bubbles come out of the solution as hydride which becomes hydrogen gas. We need to determine the ph of a 4.18 liter solution of 0.378 molar hydrochloric acid. After passing through 27.5 ampules for minutes. So we are going to recall that one. Ampule is equivalent to one column per second. So we can interpret this as actually 27.5 columns per second. Next, we want to go ahead and write out the oxidation of hydrogen gas which is formed from our hydride kati on gaining two electrons and we would bounce this out by placing a coefficient of two in front of our hydride. When we look up from our standard cell potential table, we would see for the oxidation of hydrogen gas, we have a self potential value equal to zero. Our next step is to calculate our moles of the hydride, an ion that is consumed And we would do so by taking our time given from the prompt, 39 minutes and multiplying this to convert from minutes to seconds where we recall and sorry, that should say seconds. So we would recall that we have in 60 minutes, one second. We're then going to cancel out our units of minutes and convert from units of seconds to columns by recalling that from the prompt. We have for one second, 27.5 columns, which is where we plug in that 27.5 ampule and then next, now that we can cancel seconds, we're going to convert from our Faraday's constant in columns, two moles of electrons. And we would recall that our Faraday's constant is equal to a value of 96,485 columns for one mole of an electron. And so now we're able to cancel our units of columns. And now we're going to convert from molds of our electron two moles of our hydride catalon. And we would see from our equation here that we have two moles of electrons transferred for two moles of hydride. So we would plug that in as two moles of electrons transferred for two moles of hydride. And now we're able to cancel out moles of our electrons were left with moles of hydride as our final unit. And what we're going to get here is a result equal to 0.6669 moles of hydride that is consumed. And actually just as a correction here, we're going to get a value of 1.85 times 10 to the negative fourth power moles of hydride consumed. Next, according to the prompt were given 4.18 liters and we're going to use this to find our moles of h plus initially. So we're gonna take the 4.18 L. And we're going to convert from leaders to moles by recalling that according to the solution, we're sorry, according to the prompt, we have .378 and sorry, that should be in our numerator. So we have 0.378 moles of our hydrochloric acid for one liter of solution according to the prompts. And then we're going to multiply this by our next conversion factor where we're going to go from moles of hcl, two moles of hydride and then we would recognize that we just have for one mole of hcl, one mole of hydrogen or hydride. And so now we're able to cancel out moles of hcl. We're left with moles of hydride and we're going to get a value equal to one .58004 moles of hydride initially. So now we're going to take the difference of our moles of hydride initially. So 1.58004 moles of hydride initially subtracted from our molds of hydride consumed which above we stated is 1.85 times 10 to the negative fourth power moles of hydride that is consumed. And this difference gives us a value of 1. moles of hydride remaining. And now what we're going to do is take this value remaining. So 1.57986 moles of h plus that remains. And we're going to divide this by the volume from our prompt which is given as 4.18 liters. And so in doing so we're going to get a value for our polarity equal to 0.3780 moller. And now we can determine ph by recalling that we can find our ph by taking the negative log of our concentration of H plus, which we just found above. So we would take the negative log of 0.3780 Moller. And this is going to give us our ph for a solution 0.4234, our solution and this would be our final answer since we've calculated the ph So I hope that everything I explained was clear. If you have any questions, please leave them down below. Otherwise, I'll see everyone in the next practice video.

A current of 11.3 A is applied to 1.25 L of a solution of 0.552 M HBr converting some of the H+ to H2(g), which bubles out of solution. What is the pH of the solution after 73 minutes?

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