A 10.0 mL of 0.121 M H2SO4 is neutralized by 17.1 mL of KOH...

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Transcript

00:01 Hello students let's begin with this question.

00:02 In this question some values are given.

00:05 So firstly the volume of the h2so4 is given which is equals to 10 .0 ml and then the molarity of the h2so4 is given this is equals to 0 .1 to 1 molar and then the volume of the koh is given which is equals to 17 .1 ml and in this question we have to find out the molarity of the koh.

00:35 Now firstly let's write down the balanced equation when the h2so4 reacts with the koh.

00:41 Now h2so4 this is an acid and it will react with the two moles of the potassium hydroxide this is a and this will result to the formation of the k2so4 and it will give two moles of water.

00:58 Now let's find out firstly the moles of the h2so4.

01:03 So in order to find out the moles the formula used is moles is equals to molarity multiply by volume in liters.

01:13 Now let's substitute the values over here.

01:17 So moles of h2so4 comes to be its molarity is 0 .1 to 1 molar multiply by its volume is in the question it is given in the ml.

01:30 So it is converted into the liter by dividing it by a thousand.

01:35 So here comes 0 .01 liter.

01:39 So now substitute the values over here.

01:42 So here come 0 .01 liter.

01:45 So when we calculate this we get the moles of the h2so4.

01:50 So this is equals to 0 .00121 moles...

Question

A 10.0 mL of 0.121 M $H_2SO_4$ is neutralized by 17.1 mL of KOH solution. The molarity of the KOH solution is ________. ? 0.142 M ? 0.0708 M ? 0.207 M ? 0.428 M ? 0.414 M

Question

A 10.0 mL of 0.121 M $H_2SO_4$ is neutralized by 17.1 mL of KOH solution. The molarity of the KOH solution is ________. ? 0.142 M ? 0.0708 M ? 0.207 M ? 0.428 M ? 0.414 M

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