A 20.0-mL sample of 0.25 M HNO 3 is titrated with 0.15 M NaOH. What is the pH of

Question

1. A 20.0-mL sample of 0.25 M HNO3 is titrated with 0.15 M NaOH. What is the pH of the solution after 30.0 mL of NaOH have been added to the acid? a. 2.00 b. 1.00 c. none of the answers are correct d. 1.05 e. 1.60
   
   I have seen this same question on here but I still don't understand how they got to a final answer of pH = 2.0 Could someone thoroughly show how to do this problem? Thanks.

A 20.0-mL sample of 0.25 M HNO3 is titrated with 0.15 M NaOH. What is the pH of the solution after 30.0 mL of NaOH have been added to the acid? a. 2.00 b. 1.00 c. none of the answers are correct d. 1.05 e. 1.60

I have seen this same question on here but I still don't understand how they got to a final answer of pH = 2.0 Could someone thoroughly show how to do this problem? Thanks. A 20.0-mL sample of 0.25 M HNO3 is titrated with 0.15 M NaOH. What is the pH of the solution after 30.0 mL of NaOH have been added to the acid? 2.00 1.00 none of the answers are correct 1.05 1.60

I have seen this same question on here but I still don't understand how they got to a final answer of pH = 2.0 Could someone thoroughly show how to do this problem? Thanks. a. 2.00 b. 1.00 c. none of the answers are correct d. 1.05 e. 1.60

I have seen this same question on here but I still don't understand how they got to a final answer of pH = 2.0 Could someone thoroughly show how to do this problem? Thanks.

Explanation / Answer

Moles HNO3 = 0.020 L x 0.25 =0.0050

Moles NaOH = 0.030 L x 0.15 = 0.0045

Moles HNO3 in excess = 0.0050 - 0.0045 =0.00050

total volume = 0.050 L

concentration H+ = 0.00050 / 0.050 = 0.0100 M

pH = 2.00

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