Phenolphthalein is normally used for this reaction as it accurately displays the end point of the reaction, turning colorless as soon as it reaches the end point of the reaction. If carrying out this reaction, be careful, as the change is usually quite sudden. The color exhibited should be pink (or "fuchsia") at the beginning of the reaction. As it progresses, it should become lighter in color, until it becomes colorless suddenly.
Phenolphthalein indicates red for acids and blue for bases.
This is so since the pH at the end point of Phenolphthalein is 9.1 and methyl orange is 3.7. For a strong acid strong base titration which the end point is between 3-11 phenolphthalein is used
Type your answer here... Phenolphthaelin indicator changes its colour at 8.3 pH. But the end point of this reaction is at pH 7. In weak base Strong acid case the curve gives sharp change from 3.2 to 6.0 pH. Hence phenolphthalein indicator will not be advicable.
yes it is
Acid base titration: 0.1 M HCl and phenolphtaleine indicator (pH=>8 blue)
Phenolphthalein indicates red for acids and blue for bases.
This is so since the pH at the end point of Phenolphthalein is 9.1 and methyl orange is 3.7. For a strong acid strong base titration which the end point is between 3-11 phenolphthalein is used
Type your answer here... Phenolphthaelin indicator changes its colour at 8.3 pH. But the end point of this reaction is at pH 7. In weak base Strong acid case the curve gives sharp change from 3.2 to 6.0 pH. Hence phenolphthalein indicator will not be advicable.
yes it is
Acid base titration: 0.1 M HCl and phenolphtaleine indicator (pH=>8 blue)
NaOH and HCl
What you are doing here is titration. You know you have a solution of HCl, but you do not know how much HCl is in it. For this you use something that can react with HCl (NaOH) and use an indicator to tell you when the reaction is complete. The reaction is pretty simple: HCl + NaOH --> H2O + NaCl You can see here that NaOH and HCl have a 1:1 mol relationship. So, lets find out how many moles of NaOH you used up with concentration = moles/volume 0.10 M NaOH = moles NaOH/ 0.0197 L NaOH solution Remember that M is in moles/L moles NaOH = 0.00197 moles Since you have a 1:1 relationship of NaOH with HCl, the 0.00197 mol applies to HCl as well. The next question works the same way, but backwards. Try doing it yourself if you understood the first part before reading my answer. Find out how many moles of HCl you have so you can find out how much moles of NaOH you need for the neutralization. 0.050 M HCl = n HCl / 0.020 L HCl soln n HCl = 0.001 mol HCl Remeber the 1:1 relationship, which gives you that n NaOH = 0.001 mol. Now all you need is the volume. 0.1 M NaOH = 0.001 mol NaOH/ Volume soln V = 0.01 L = 10 mL
I don't know ... I have the same problem . I found this in a chemestry problem , and my teacher couldn't tell me the answer . He told me to search for it, but I can't find anything about the reaction between Methyl Orange and HCl .. If you could help me , please leave a message at YM : valentin_gr2008@yahoo.com Thanks !
iv done the titration and the preferred indicator for such a solution is phenalphaline, use a few drops to make it go a bright luminous pink and titrate till clear, not 'cloudy'
For titration of hci in a mixture of hcl and acetic acid, the indicator used is
we know that hcl is an acid and acids change purple phenolphthalein to colourless. So HCL changes purple phenolphthalein to colourless
The product is sodium chloride.The reaction is:NaOH + HCl - NaCl + H2O