Suppose that 19.46 mL of standardized 0.0941 M NaOH is...

Suppose that 19.46 mL of standardized 0.0941 M NaOH is required to neutralize 15.00 mL of H2SO4 solution. a. Write the balanced chemical equation. b. Calculate the molarity of the sulfuric acid solution. A student finds that 0.010 L of H3PO4 requires 8.15 mL of 3.4 x 10^-3 M NaOH to neutralize. a. Write the balanced chemical equation. b. Calculate the molarity of the phosphoric acid solution. It is determined that 0.0364 L of 7.6 x 10^-2 M Ba(OH)2 solution is needed to neutralize 13.5 mL of hydrochloric acid (HCl) solution of unknown concentration a. Write the balanced chemical equation. b. What is the molarity of the HCl? A procedure calls for 0.0850 M KOH to neutralize 8.00 mL of 0.125 M H3PO4. a. Write the balanced chemical equation. b. How many milliliters of KOH will be required? A solution of 0.721 M HNO3 can be used to neutralize 0.0175 L of 0.261 M NaOH. a. Write the balanced chemical equation. b. How many milliliters of HNO3 will be required? A solution of 0.412 M LiOH is used to neutralize 44.5 mL of 0.0105 M H2SO4. a. Write the balanced chemical equation. b. How many milliliters of LiOH will be required?

Transcript

00:01 Hello students, so in this video has six parts.

00:02 In the first part, we have to balance the equation naoh plus h2so4 giving rise to na2so4 plus water.

00:12 So firstly, we will balance the na and then we will balance the hydrogen.

00:15 So we have 4 here.

00:16 So to balance it, we will write here 2.

00:17 So this is now balanced.

00:18 Now we have 19 .46 ml of naoh and 0 .0941 molar of naoh and we have 15 ml of h2so4.

00:27 We don't know the molarity.

00:28 So now the molarity can be calculated using this equation mnaoh into vnaoh into nh2so4 is equals to mh2so4 into vh2so4 into nnaoh.

00:44 So that is equals to 0 .0941 into 19 .46 into 1 is equals to mh2so4 into 15 into 2.

00:57 So mh2so4 is equals to 0 .0941 into 19 .46 by 15 into 2 that is equals to 0 .061 molar.

01:08 Let's move on to the next.

01:09 Now for the second part, we have h3po4 reacting with naoh to form na3po4 plus water.

01:19 So again, we will balance the na plus first and then we have to balance the hydrogen.

01:24 So we have 4 hydrogen here.

01:26 So we'll write here 2 and it is balanced now.

01:28 So we have 0 .010 liter and we don't know the molarity of h3po4 and we have 8 .15 ml of naoh and 3 .4 into 10 to the power minus 3 molar of naoh.

01:38 So we know that mnaoh into vnaoh into nh3po4 is equals to mh3po4 into vh3po4 into nnaoh that is equals to 3 .4 into 10 to the power minus 3 into 8 .15 into, so we will convert this to liter into 1 is equals to mh3po4 into the volume is 0 .010 and it is 3 that is equals to mh3po4 is equals to 3 .4 into 10 to the power minus 3 into 8 .15 into 10 to the power minus 3 divided by 0 .010 into 3 that is equals to 9 .24 into 10 to the power minus 4 molar.

02:33 Let's move on to the next one.

02:34 Now for the next one, we have barium hydroxide reacting with hcl to produce bacl2 plus h2o.

02:42 Now we will again balance the cl first and then hydrogen.

02:46 So we have 0 .0364 liter and 7 .6 into 10 to the power minus 2 molar of barium hydroxide and we have 13 .5 ml of hcl.

02:56 We have to find the molarity.

02:57 So we know that mbaoh whole twice into vbaoh whole twice into n of hcl is equals to m of hcl into v of hcl into n of baoh whole twice that is equals to 7 .6 into 10 to the power minus 2 into 0 .0364 into 2 is equals to mhcl into 13 .5 into so we will convert this to liters into 1 that is equals to mhcl is equals to 0 .0364 into 7 .6 into 10 to the power minus 2 into 2 into 1000 divided by 13 .5 that is equals to 0 .409 molar.

03:49 Let's move on to the next one...

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