In an acid base titration 32.5 ml of sodium hydroxide were neautralized by 17.6 ml of 0.180 mol/l sulfuric acid. Calculate the concentration of the sodium hydroxide.
I asked this question earlier but i wanted to make sure i got the right answer
My answer: 0.36 mol/L is this correct
I don't think so. Did I post the wrong instructions? If you will post your work I will look at it.
Ok.
Concentration NaOH = (32.5 ml)(0.180 mol/L H2SO4)(2NaOH/H2SO4) divided by 32.5 ml
DrBob222
I should my work
Can you please verify if i did it correctly, as you said my answer is wrong
The problem states 17.5 mL H2SO4 and not 32.5. The 32.5 mL is volume NaOH in the problem.
If I replace the first 32.5 with 17.6 your set up is right. Then multiply 17.6 x 0.180 x 2/32.5 I don't get 0.36 but I do get the right answer. You may be punching it in on the calculator wrong.
Ohhhh is it 0.19
Almost. It's 0.19?. Don't throw that last digit away. You have three places in the volume and three places in the molarity; therefore, you're allowed three places in the answer. You're throwing a perfectly good number so you should keep it. I have 0.1949538 which I would round to 0.195
Can you also verify my answer for a question i asked hours ago..
Yes. Can you give me a link? I don't think you can post a link so look at the link and give me the time and date in the line. I can find it that way. Also the name/author.
Name: Jack
Date: Monday January 12, 2015
Time: 5:28 pm
The amswer i got was 0.257 L