HCOOH + H2O <---> HCOO- + H3O+
<span>Ka = 1.8 X 10^-4 = [HCOO-][H3O+] / [HCOOH] </span>
<span>In the solution, [HCOO-] = [H3O+] = x </span>
<span>Quite properly, [HCOOH] = (2.30 X 10^-3 - x). </span>
<span>Since the formic acid solution is pretty dilute, and since Ka is sort of small, we can initially assume that x will be small compared to 2.3 X 10^-3, and so we can ignore it. If we do that, then, </span>
<span>Ka = x^2 / 2.30X10^-3 = 1.80 X 10^-4 </span>
<span>x = 6.4 X 10^-4 = [H3O+] </span>
<span>pH = 3.19 </span>
<span>Now, quite properly, our assumption that x would be small compared to 2.3 X 10^-3 is incorrect, and we really cannot ignore x in that expression. So, we should go back to the original expression for Ka: </span>
<span>Ka = x^2 / (2.30 X 10^-3 - x) = 1.80 X 10^-4 </span>
<span>Quite properly, you should rearrange this into a quadratic form and use the quadratic equation to solve for x. Once you've done that, x = [H3O+], and pH = - log (x).</span>
<span>1 gallon = 4. 5 litres
1 litre = 1/4.5 gallons = 2/9 gallons
1 km = 5/8 mile
9.8 km = 9.8 x 5/8 miles = 49/8 miles
49/8
-------- mpg
2/9
49 x 9
--------- mpg
16
27.6 mpg </span>
The subduction's zones boundaries are the biggest collision between two tectonic plates.
Answer:
0.9
Explanation:
The pka represents the force by which the molecules need to dissociate for the acids ,
Hence , lower the pka stronger will be the acid and that therefore will dissociate completely and vice versa , for a weak acid higher the pka .
And in case of a base , it will be completely reversed , lower pKa , weaker base ,
and higher pKa , stronger base .
From the data of the question ,
0.9 is the lowest value of the pKa , hence , weakest base .
Answer:
High temperatures
Explanation:
NaHCO₃ (8) + HCH,O₂ (aq) → H₂O (l) + CO₂ (g) + NaC,H₃O₂ (aq)
As the flask gets cooler to the touch as the reaction proceeds, the reaction is endothermic. This means that ΔH is positive (ΔH>0).
As a gas is formed (bubbles are formed), ΔS is positive (ΔS>0).
<em>In terms of ΔG:</em>
<em>In order for the reaction to be thermodynamically favorable, ΔH has to be negative</em>, thus:
- The reaction is favorable if TΔS > ΔH.
The greater the temperature, the easier it would be for TΔS to be greater than ΔH.