Asked by Sandy

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
Answered by DrBob222
I don't think so. Did I post the wrong instructions? If you will post your work I will look at it.
Answered by Sandy
Ok.
Concentration NaOH = (32.5 ml)(0.180 mol/L H2SO4)(2NaOH/H2SO4) divided by 32.5 ml
Answered by Sandy
DrBob222
I should my work
Can you please verify if i did it correctly, as you said my answer is wrong
Answered by DrBob222
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.
Answered by Sandy
Ohhhh is it 0.19
Answered by DrBob222
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
Answered by Sandy
Can you also verify my answer for a question i asked hours ago..
Answered by DrBob222
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.
Answered by Sandy
Name: Jack
Date: Monday January 12, 2015
Time: 5:28 pm
The amswer i got was 0.257 L
Answered by DrBob222
ok. I found it and I'm copying it here. However, I don't see an answer.
Answered by Sandy
I posted my answer here in this link
Answered by Sandy
When i gave you the name and time i wrote my answer
Answered by DrBob222
And I didn't get the link posted but here it is.
http://www.jiskha.com/display.cgi?id=1421101733

I get almost 257 but the number if have is 0.25644 L which I would round to 0.256. I don't know what atomic mass you used for Ca. I went back and checked and used 40.078 and that gave me a number of 0.256589 L. I would round that to 0.257 also. Good work. I used 40.1 to get the 0.256 above.
Answered by Sandy
Thank you soo much DrBob222
You helped me a lot.
Thank you very much. XD
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