Part C 0.180 M in HCHO2 (Ka = 1.8 × 10^-4) and 0.230 M in...

Part C 0.180 M in HCHO2 ($K_a = 1.8 \times 10^{-4}$) and 0.230 M in HC2H3O2 ($K_a = 1.8 \times 10^{-5}$) Express your answer to two decimal places. pH = 2.09 Submit Previous Answers Request Answer Incorrect; Try Again; 3 attempts remaining Part D 5.00$\times$10$^{-2}$ M in acetic acid ($K_a = 1.8 \times 10^{-5}$) and 5.00$\times$10$^{-2}$ M in hydrocyanic acid ($K_a = 4.9 \times 10^{-10}$) Express your answer to two decimal places. pH = Submit Request Answer Provide Feedback

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00:01 For the given question we are given with the data of the ch3cooh as the 0 .015 molar and the hcn as the 0 .015 molar and we have to find the combined ph.

00:17 How we will find combined ph? by equating both h positive ions concentration.

00:29 This we can do for finding the combined ph.

00:33 Firstly, we will draw the ice table for the ch3cooh in the aqueous medium will give dissociation with ch3cooo and h positive in the aqueous medium.

00:48 So at the initial condition it will be 0 .015 no reaction is happening therefore 0 and 0 but at equilibrium some constant will be subtracted from 0 .015 as the reaction takes place and product is formed therefore it will be added as the product for the x and x will be the constant added to it.

01:13 So ka value is given to us as 1 .8 multiplied by 10 to the power minus 5 and ka means the formula product upon the reactant.

01:26 This is the formula of the k.

01:28 So we will equate 1 .8 multiplied by 10 to the power minus 5 with the x and x divided by the 1 0 .015 minus x but this denominator x is very negligible therefore we will not in the consideration for this x and we will neglect it.

01:50 Therefore it will be equal to 1 .8 multiplied by 10 to the power minus 5 multiplied by the 0 .015 will be equal to the x square.

01:59 So x square will be equal to the 0 .27 multiplied by 10 to the power minus 6.

02:05 So x will be equal to the 0 .5196 multiplied by 10 to the power minus 3 and this is also the concentration of the h positive ion...

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