Calculate the pH of a 5.942 mL solution containing 1.1 g of nitric acid
5 answers
actually, it's 5,942 mL not 5.942
Is it 2.53?
I don't think so. Post your work so I can find the error and know what's going on.
I used an online calculator. Our teacher didn't explain this well on how to work pH
moles HNO3 = 1g/63 = 0.0159
M HNO3 = moles/L = 0.0159/5.942 = 0.00267. HNO3 is a strong acid; therefore, this is the (H^+). Then pH = -log(0.00267) = 2.57
Again, on-line calculators are not always available. You should learn to do these yourself. Look at my solution and follow up with whatever questions you have about how to do it yourself.
M HNO3 = moles/L = 0.0159/5.942 = 0.00267. HNO3 is a strong acid; therefore, this is the (H^+). Then pH = -log(0.00267) = 2.57
Again, on-line calculators are not always available. You should learn to do these yourself. Look at my solution and follow up with whatever questions you have about how to do it yourself.
1 answer